Номер 524, страница 245 - гдз по алгебре 10-11 класс учебник Колмогоров, Абрамов
Авторы: Колмогоров А. Н., Абрамов А. М., Дудницын Ю. П.
Тип: Учебник
Издательство: Просвещение
Год издания: 2008 - 2026
Цвет обложки: зелёный, чёрный
ISBN: 978-5-09-019513-3
Рекомендовано Министерством образования и науки Российской Федерации
Алгебра и начала математического анализа
Популярные ГДЗ в 10 классе
Глава 4. Показательная и логарифмическая функции. Параграф 10. Показательная и логарифмическая функции - номер 524, страница 245.
App\Models\Task {#1030 // resources/views/models/task/default.blade.php #connection: "mysql" #table: "tasks" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:24 [ "id" => 104355 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_page_start" => "245" "field_page_end" => null "field_url" => "/10-klass/algebra/kolmogorov/524" "field_display_title" => "524" "field_outside_task" => null "field_task_type" => Illuminate\Database\Eloquent\Collection {#1037 #items: array:1 [ 0 => App\Models\Term {#1036 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 26 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "номер" "field_cases" => array:6 [ "field_accusative_case" => "номер" "field_creative_case" => "номером" "field_dative_case" => "номеру" "field_genitive_case" => "номера" "field_nominative_case" => "номер" "field_prepositional_case" => "номере" ] "field_short_name" => "№" ] #original: array:6 [ "id" => 26 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "номер" "field_cases" => array:6 [ "field_accusative_case" => "номер" "field_creative_case" => "номером" "field_dative_case" => "номеру" "field_genitive_case" => "номера" "field_nominative_case" => "номер" "field_prepositional_case" => "номере" ] "field_short_name" => "№" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_match" => null "breadcrumbs" => [] "edition_groups" => Illuminate\Database\Eloquent\Collection {#1035 #items: [] #escapeWhenCastingToString: false } "top_parent_branch" => Illuminate\Database\Eloquent\Collection {#1046 #items: array:1 [ 0 => App\Models\Branch {#1045 #connection: "mysql" #table: "branches" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:24 [ "id" => 103776 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Показательная и логарифмическая функции" "field_branch_order" => "4" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1050 #items: array:1 [ 0 => App\Models\Term {#1047 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#observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_publisher" => Illuminate\Database\Eloquent\Collection {#1055 #items: array:1 [ 0 => App\Models\Term {#1056 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 5153 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Просвещение" "field_cases" => null "field_translit" => "prosveschenie" ] #original: array:6 [ "id" => 5153 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Просвещение" "field_cases" => null "field_translit" => "prosveschenie" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] 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"name" => "Москва" "field_cases" => array:6 [ …6] "field_translit" => "moskva" ] #original: array:6 [ "id" => 15 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Москва" "field_cases" => array:6 [ …6] "field_translit" => "moskva" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_series" => Illuminate\Database\Eloquent\Collection {#1068 #items: [] #escapeWhenCastingToString: false } "field_umk" => Illuminate\Database\Eloquent\Collection {#1069 #items: [] #escapeWhenCastingToString: false } "field_level_of_education" => Illuminate\Database\Eloquent\Collection {#1070 #items: [] #escapeWhenCastingToString: false } "field_standart_of_education" => Illuminate\Database\Eloquent\Collection {#1071 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+usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_under_the_edition" => Illuminate\Database\Eloquent\Collection {#1074 #items: array:1 [ 0 => App\Models\Term {#1058} ] #escapeWhenCastingToString: false } "field_under_the_edition_degree" => null "field_cover_description" => null "field_publication_year" => "2008" "field_publication_year_until" => null "field_part" => "" "field_part_writing" => "" "field_second_foreign_language" => null "field_for_whom" => "Учебник для 10-11 классов общеобразовательных учреждений" "field_allowed" => "Рекомендовано Министерством образования и науки Российской Федерации" "field_reserve_field" => "Алгебра и начала математического анализа" "field_link_to_source" => null "field_tasks_count" => "1202" "field_priority" => "2" "field_default_folder" => "/algebra_10/kolmogorov/" "field_isbn" => "978-5-09-019513-3" "field_cover" => array:1 [ 0 => 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=> "2026-04-10 13:58:26" "updated_at" => null "name" => "чёрный" "field_cases" => array:6 [ …6] "field_translit" => "chyornyy" ] #original: array:6 [ "id" => 56 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "чёрный" "field_cases" => array:6 [ …6] "field_translit" => "chyornyy" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => "<!--Текст в низу страницы--> <p> Выбирая в качестве онлайн-консультанта <strong>ГДЗ по алгебре 10-11 класс учебник Колмогоров, Абрамов, Дудницын (Просвещение)</strong>, каждый школьник может быть уверен, что справится со всеми учебными задачами текущего года. Помимо идеальной подготовки домашних заданий, подросток сможет самостоятельно осваивать новые темы, выполнять диагностику своих знаний и восполнять в них пробелы, повторять и закреплять изученный материал. При этом ребенку не понадобятся дополнительные консультации педагога. Со всеми вопросами и темами текущего учебного года он справится самостоятельно. </p> <h2> Что готовит школьникам алгебра в 10-11 классах </h2> <p> Алгебру можно назвать лидером среди всех учебных дисциплин по количеству изучаемых тем. В десятом и одиннадцатом классах все усложняется тем, что помимо новых разделов, ребятам предстоит огромная работа по повторению ранее пройденного материала, ведь в конце года всех их ждет сложнейшее испытание - итоговый экзамен по предмету. Но его результат будет зависеть от того, насколько хорошо ребята освоят следующие разделы: </p> <ol> <li>Тригонометрические функции.</li> <li>Производная и её применения.</li> <li>Первообразная и интеграл.</li> <li>Показательная и логарифмическая функции.</li> <li>Действительные числа.</li> <li>Функции. </ol> <p> Материал очень насыщенный и достаточно сложный. При этом недостаточно будет просто вызубрить массу формул и правил. Простое заучивание - это лишь часть работы. Главное научиться применять полученные знания на практике. Именно с этим у большинства ребят возникают проблемы. Работа вместе с <strong>решебником по алгебре за 10-11 классы авторов Колмогорова, Абрамова, Дудницына</strong> позволит преодолеть все препятствия в учебе. Детально расписанная в пособии информация поможет прекрасно понять материал и научиться применять теорию на практике. </p> <h2> Что представляет собой решебник к учебнику Колмогорова (Просвещение) </h2> <p> Верные ответы — это далеко не все, что содержится во вспомогательном пособии. Сборник с готовыми домашними заданиями структурно и по содержанию полностью копирует учебник под редакцией Колмогорова и включает: </p> <ul> <li>тематические параграфы с соответствующими им упражнениями;</li> <li>номера к разделу «Задачи на повторение»;</li> <li>задачи повышенной трудности.</li> </ul> <p> Каждое решение расписано максимально подробно, с выкладками по теории, графиками и выводами. В особо сложных вопросах авторы дают дополнительные комментарии. Материал изложен простым языком, доступным для понимания абсолютно всем старшеклассникам. Электронный формат существенно облегчает работу с пособием, позволяет выбрать удобный ритм для учебы, экономит время и силы ребенка. </p> <h2> Зачем использовать ГДЗ по алгебре </h2> <p> Даже в выпускном классе нельзя опускать руки, если вдруг освоение алгебры не задалось. Еще есть время поправить свои оценки и знания. Главное правильно распределить свое время, найти верного помощника и вернуть интерес к предмету. Решебник позволит все эти элементы совместить воедино для достижения блестящего результата. Предложенное пособие станет тем консультантом, который даст возможность подростку быстро и качественно выполнить домашнее задание. Это в свою очередь позволит вернуть интерес к предмету и, что немало важно, выкроить ценные часы на отдых. Главное - не прибегать к списыванию готовых решений, ведь простое копирование поможет лишь на время, однако на первой же проверочной в классе обман вскроется. 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Помимо идеальной подготовки домашних заданий, подросток сможет самостоятельно осваивать новые темы, выполнять диагностику своих знаний и восполнять в них пробелы, повторять и закреплять изученный материал. При этом ребенку не понадобятся дополнительные консультации педагога. Со всеми вопросами и темами текущего учебного года он справится самостоятельно. </p> <h2> Что готовит школьникам алгебра в 10-11 классах </h2> <p> Алгебру можно назвать лидером среди всех учебных дисциплин по количеству изучаемых тем. В десятом и одиннадцатом классах все усложняется тем, что помимо новых разделов, ребятам предстоит огромная работа по повторению ранее пройденного материала, ведь в конце года всех их ждет сложнейшее испытание - итоговый экзамен по предмету. 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#fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "content" => Illuminate\Database\Eloquent\Collection {#1112 #items: array:5 [ 0 => App\Models\Element {#1103 #connection: "mysql" #table: "elements" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:7 [ "id" => 156767 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1096 #items: array:1 [ 0 => App\Models\Edition {#1104 #connection: "mysql" #table: "editions" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 644 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Условие" "field_order" => "1" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1101 …2} "field_content_type" => "free" "field_content_mode" => "text, image" "field_page_content_mode" => "image" "field_content_text_checked" => null "field_page_content_text_checked" => "0" "field_solution_author" => "Автор" "field_moderator" => "vadim" "field_edition_type" => "statement" "field_root_dir" => "0-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1100 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1099 …2} "field_process_formula" => "katex" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1097 …2} "field_content_source" => null ] #original: array:21 [ "id" => 644 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Условие" "field_order" => "1" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1101 …2} "field_content_type" => "free" "field_content_mode" => "text, image" "field_page_content_mode" => "image" "field_content_text_checked" => null "field_page_content_text_checked" => "0" "field_solution_author" => "Автор" "field_moderator" => "vadim" "field_edition_type" => "statement" "field_root_dir" => "0-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1100 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1099 …2} "field_process_formula" => "katex" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1097 …2} "field_content_source" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "task" => array:2 [ "refs" => "104355" "type" => "task" ] "text" => "<p><strong>524.</strong>— <strong>a)</strong> $log_2 (9 - 2^x) = 3 - x;$</p><p><strong>б)</strong> $log_2 (25^{x+3} - 1) = 2 + log_2 (5^{x+3} + 1);$</p><p><strong>в)</strong> $log_4 (2 \cdot 4^{x-2} - 1) = 2x - 4;$</p><p><strong>г)</strong> $log_2 (4^x + 4) = log_2 2^x + log_2 (2^{x+1} - 3).$</p>" "img" => array:1 [ 0 => array:5 [ "name" => "524-1.jpg" "alt" => null "width" => "1035" "height" => 286 "path" => "/media/algebra_10/kolmogorov/0-00/524-1.webp?ts=1733865066" ] ] ] #original: array:7 [ "id" => 156767 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1096} "task" => array:2 [ "refs" => "104355" "type" => "task" ] "text" => "<p><strong>524.</strong>— <strong>a)</strong> $log_2 (9 - 2^x) = 3 - x;$</p><p><strong>б)</strong> $log_2 (25^{x+3} - 1) = 2 + log_2 (5^{x+3} + 1);$</p><p><strong>в)</strong> $log_4 (2 \cdot 4^{x-2} - 1) = 2x - 4;$</p><p><strong>г)</strong> $log_2 (4^x + 4) = log_2 2^x + log_2 (2^{x+1} - 3).$</p>" "img" => array:1 [ 0 => array:5 [ "name" => "524-1.jpg" "alt" => null "width" => "1035" "height" => 286 "path" => "/media/algebra_10/kolmogorov/0-00/524-1.webp?ts=1733865066" ] ] ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 1 => App\Models\Element {#1098 #connection: "mysql" #table: "elements" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 157654 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1087 #items: array:1 [ 0 => App\Models\Edition {#1095 #connection: "mysql" #table: "editions" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 645 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 1" "field_order" => "2" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1094 …2} "field_content_type" => "free" "field_content_mode" => "image" "field_page_content_mode" => "" "field_content_text_checked" => "0" "field_page_content_text_checked" => "0" "field_solution_author" => "Неизвестный" "field_moderator" => "vadim" "field_edition_type" => "solution" "field_root_dir" => "1-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1092 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1088 …2} "field_process_formula" => "" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1091 …2} "field_content_source" => null ] #original: array:21 [ "id" => 645 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 1" "field_order" => "2" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1094 …2} "field_content_type" => "free" "field_content_mode" => "image" "field_page_content_mode" => "" "field_content_text_checked" => "0" "field_page_content_text_checked" => "0" "field_solution_author" => "Неизвестный" "field_moderator" => "vadim" "field_edition_type" => "solution" "field_root_dir" => "1-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1092 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1088 …2} "field_process_formula" => "" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1091 …2} "field_content_source" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "task" => array:2 [ "refs" => "104355" "type" => "task" ] "img" => array:2 [ 0 => array:5 [ "name" => "524-1.png" "alt" => null "width" => "740" "height" => 1192 "path" => "/media/algebra_10/kolmogorov/1-00/524-1.webp?ts=1733865352" ] 1 => array:5 [ "name" => "524-2.png" "alt" => null "width" => "733" "height" => 577 "path" => "/media/algebra_10/kolmogorov/1-00/524-2.webp?ts=1733865352" ] ] ] #original: array:6 [ "id" => 157654 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1087} "task" => array:2 [ "refs" => "104355" "type" => "task" ] "img" => array:2 [ 0 => array:5 [ "name" => "524-1.png" "alt" => null "width" => "740" "height" => 1192 "path" => "/media/algebra_10/kolmogorov/1-00/524-1.webp?ts=1733865352" ] 1 => array:5 [ "name" => "524-2.png" "alt" => null "width" => "733" "height" => 577 "path" => "/media/algebra_10/kolmogorov/1-00/524-2.webp?ts=1733865352" ] ] ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 2 => App\Models\Element {#1090 #connection: "mysql" #table: "elements" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 158361 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1114 #items: array:1 [ 0 => App\Models\Edition {#1086 #connection: "mysql" #table: "editions" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 646 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 2" "field_order" => "3" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1085 …2} "field_content_type" => "free" "field_content_mode" => "image" "field_page_content_mode" => "" "field_content_text_checked" => "0" "field_page_content_text_checked" => "0" "field_solution_author" => "Неизвестный" "field_moderator" => "vadim" "field_edition_type" => "solution" "field_root_dir" => "2-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1084 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1043 …2} "field_process_formula" => "" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1041 …2} "field_content_source" => null ] #original: array:21 [ "id" => 646 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 2" "field_order" => "3" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1085 …2} "field_content_type" => "free" "field_content_mode" => "image" "field_page_content_mode" => "" "field_content_text_checked" => "0" "field_page_content_text_checked" => "0" "field_solution_author" => "Неизвестный" "field_moderator" => "vadim" "field_edition_type" => "solution" "field_root_dir" => "2-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1084 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1043 …2} "field_process_formula" => "" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1041 …2} "field_content_source" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "task" => array:2 [ "refs" => "104355" "type" => "task" ] "img" => array:1 [ 0 => array:5 [ "name" => "524-1.jpg" "alt" => null "width" => "1121" "height" => 740 "path" => "/media/algebra_10/kolmogorov/2-00/524-1.webp?ts=1733996022" ] ] ] #original: array:6 [ "id" => 158361 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1114} "task" => array:2 [ "refs" => "104355" "type" => "task" ] "img" => array:1 [ 0 => array:5 [ "name" => "524-1.jpg" "alt" => null "width" => "1121" "height" => 740 "path" => "/media/algebra_10/kolmogorov/2-00/524-1.webp?ts=1733996022" ] ] ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 3 => App\Models\Element {#1034 #connection: "mysql" #table: "elements" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 168651 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1133 #items: array:1 [ 0 => App\Models\Edition {#1039 #connection: "mysql" #table: "editions" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 647 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 3" "field_order" => "4" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1122 …2} "field_content_type" => "free" "field_content_mode" => "image" "field_page_content_mode" => "" "field_content_text_checked" => "0" "field_page_content_text_checked" => "0" "field_solution_author" => "Неизвестный" "field_moderator" => "vadim" "field_edition_type" => "solution" "field_root_dir" => "3-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1118 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1116 …2} "field_process_formula" => "" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1119 …2} "field_content_source" => null ] #original: array:21 [ "id" => 647 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 3" "field_order" => "4" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1122 …2} "field_content_type" => "free" "field_content_mode" => "image" "field_page_content_mode" => "" "field_content_text_checked" => "0" "field_page_content_text_checked" => "0" "field_solution_author" => "Неизвестный" "field_moderator" => "vadim" "field_edition_type" => "solution" "field_root_dir" => "3-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1118 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1116 …2} "field_process_formula" => "" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1119 …2} "field_content_source" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "task" => array:2 [ "refs" => "104355" "type" => "task" ] "img" => array:1 [ 0 => array:5 [ "name" => "524-1.jpg" "alt" => null "width" => "1121" "height" => 740 "path" => "/media/algebra_10/kolmogorov/3-00/524-1.webp?ts=1733995977" ] ] ] #original: array:6 [ "id" => 168651 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1133} "task" => array:2 [ "refs" => "104355" "type" => "task" ] "img" => array:1 [ 0 => array:5 [ "name" => "524-1.jpg" "alt" => null "width" => "1121" "height" => 740 "path" => "/media/algebra_10/kolmogorov/3-00/524-1.webp?ts=1733995977" ] ] ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 4 => App\Models\Element {#1117 #connection: "mysql" #table: "elements" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 1342178 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1141 #items: array:1 [ 0 => App\Models\Edition {#1132 #connection: "mysql" #table: "editions" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 5114 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 5" "field_order" => "6" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1135 …2} "field_content_type" => "free" "field_content_mode" => "text" "field_page_content_mode" => "" "field_content_text_checked" => null "field_page_content_text_checked" => "0" "field_solution_author" => "Gemini 2.5 Pro" "field_moderator" => "vadim" "field_edition_type" => "solution" "field_root_dir" => "5-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1136 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1137 …2} "field_process_formula" => "katex" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1138 …2} "field_content_source" => null ] #original: array:21 [ "id" => 5114 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 5" "field_order" => "6" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1135 …2} "field_content_type" => "free" "field_content_mode" => "text" "field_page_content_mode" => "" "field_content_text_checked" => null "field_page_content_text_checked" => "0" "field_solution_author" => "Gemini 2.5 Pro" "field_moderator" => "vadim" "field_edition_type" => "solution" "field_root_dir" => "5-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1136 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1137 …2} "field_process_formula" => "katex" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1138 …2} "field_content_source" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "task" => array:2 [ "refs" => "104355" "type" => "task" ] "text" => "<p><strong>а)</strong> $log_2(9 - 2^x) = 3 - x$</p><p>Определим область допустимых значений (ОДЗ). Аргумент логарифма должен быть строго положительным:<br>$9 - 2^x > 0$<br>$9 > 2^x$<br>$log_2(9) > x$</p><p>Теперь решим уравнение. По определению логарифма ($log_a b = c \Leftrightarrow a^c = b$):<br>$9 - 2^x = 2^{3-x}$<br>$9 - 2^x = \frac{2^3}{2^x}$<br>$9 - 2^x = \frac{8}{2^x}$</p><p>Сделаем замену переменной. Пусть $t = 2^x$. Так как $2^x > 0$ для любого $x$, то $t > 0$.<br>$9 - t = \frac{8}{t}$<br>Умножим обе части на $t$ (так как $t \neq 0$):<br>$t(9 - t) = 8$<br>$9t - t^2 = 8$<br>$t^2 - 9t + 8 = 0$</p><p>Это квадратное уравнение. Решим его с помощью теоремы Виета или через дискриминант.<br>Корни уравнения: $t_1 = 1$, $t_2 = 8$. Оба корня положительны, поэтому оба подходят.</p><p>Вернемся к исходной переменной $x$:<br>1) $2^x = t_1 = 1 \Rightarrow 2^x = 2^0 \Rightarrow x_1 = 0$.<br>2) $2^x = t_2 = 8 \Rightarrow 2^x = 2^3 \Rightarrow x_2 = 3$.</p><p>Проверим, удовлетворяют ли найденные корни ОДЗ ($x < log_2(9)$).<br>$log_2(9) \approx 3.17$.<br>1) $x_1 = 0$. $0 < log_2(9)$, корень подходит.<br>2) $x_2 = 3$. $3 < log_2(9)$, так как $log_2(8) < log_2(9)$, корень подходит.</p><p>Ответ: $0; 3$.</p><p><strong>б)</strong> $log_2(25^x + 3 - 1) = 2 + log_2(5^x + 3 + 1)$</p><p>Упростим выражения в скобках:<br>$log_2(25^x + 2) = 2 + log_2(5^x + 4)$</p><p>ОДЗ:<br>1) $25^x + 2 > 0$. Это выражение всегда положительно, так как $25^x > 0$.<br>2) $5^x + 4 > 0$. Это выражение также всегда положительно, так как $5^x > 0$.<br>Следовательно, ОДЗ: $x \in \mathbb{R}$.</p><p>Преобразуем уравнение, используя свойства логарифмов:<br>$log_2(25^x + 2) - log_2(5^x + 4) = 2$<br>$log_2\left(\frac{25^x + 2}{5^x + 4}\right) = 2$</p><p>По определению логарифма:<br>$\frac{25^x + 2}{5^x + 4} = 2^2$<br>$\frac{25^x + 2}{5^x + 4} = 4$</p><p>Сделаем замену переменной. Пусть $t = 5^x$, тогда $25^x = (5^x)^2 = t^2$. Условие $t > 0$.<br>$\frac{t^2 + 2}{t + 4} = 4$<br>$t^2 + 2 = 4(t + 4)$<br>$t^2 + 2 = 4t + 16$<br>$t^2 - 4t - 14 = 0$</p><p>Решим квадратное уравнение через дискриминант:<br>$D = (-4)^2 - 4(1)(-14) = 16 + 56 = 72$<br>$t = \frac{4 \pm \sqrt{72}}{2} = \frac{4 \pm 6\sqrt{2}}{2} = 2 \pm 3\sqrt{2}$<br>Получаем два корня: $t_1 = 2 + 3\sqrt{2}$ и $t_2 = 2 - 3\sqrt{2}$.</p><p>Так как $t = 5^x$, то $t$ должно быть положительным. $t_1 = 2 + 3\sqrt{2} > 0$.<br>$t_2 = 2 - 3\sqrt{2} \approx 2 - 3 \cdot 1.41 = 2 - 4.23 < 0$. Этот корень не подходит.</p><p>Вернемся к замене: $5^x = 2 + 3\sqrt{2}$<br>$x = log_5(2 + 3\sqrt{2})$</p><p>Ответ: $log_5(2 + 3\sqrt{2})$.</p><p><strong>в)</strong> $log_4(2 \cdot 4^{x-2} - 1) = 2x - 4$</p><p>ОДЗ:<br>$2 \cdot 4^{x-2} - 1 > 0$<br>$2 \cdot \frac{4^x}{4^2} - 1 > 0$<br>$2 \cdot \frac{4^x}{16} - 1 > 0$<br>$\frac{4^x}{8} > 1$<br>$4^x > 8$<br>$(2^2)^x > 2^3 \Rightarrow 2^{2x} > 2^3 \Rightarrow 2x > 3 \Rightarrow x > \frac{3}{2}$</p><p>Решим уравнение, используя определение логарифма:<br>$2 \cdot 4^{x-2} - 1 = 4^{2x-4}$<br>Заметим, что $2x-4 = 2(x-2)$. Тогда $4^{2x-4} = 4^{2(x-2)} = (4^{x-2})^2$.</p><p>Сделаем замену $t = 4^{x-2}$. Уравнение примет вид:<br>$2t - 1 = t^2$<br>$t^2 - 2t + 1 = 0$<br>$(t - 1)^2 = 0$<br>$t = 1$</p><p>Вернемся к замене:<br>$4^{x-2} = 1$<br>$4^{x-2} = 4^0$<br>$x - 2 = 0$<br>$x = 2$</p><p>Проверим корень по ОДЗ ($x > 3/2$).<br>$2 > 1.5$, корень подходит.</p><p>Ответ: $2$.</p><p><strong>г)</strong> $log_2(4^x + 4) = log_2(2^x) + log_2(2^{x+1} - 3)$</p><p>ОДЗ:<br>1) $4^x + 4 > 0$ (верно для всех $x$).<br>2) $2^x > 0$ (верно для всех $x$).<br>3) $2^{x+1} - 3 > 0 \Rightarrow 2 \cdot 2^x > 3 \Rightarrow 2^x > \frac{3}{2} \Rightarrow x > log_2(\frac{3}{2})$.<br>Общее ОДЗ: $x > log_2(1.5)$.</p><p>Используем свойство суммы логарифмов $log_a b + log_a c = log_a(bc)$:<br>$log_2(4^x + 4) = log_2(2^x \cdot (2^{x+1} - 3))$</p><p>Так как основания логарифмов равны, приравниваем их аргументы:<br>$4^x + 4 = 2^x(2 \cdot 2^x - 3)$</p><p>Сделаем замену $t = 2^x$. Тогда $4^x = t^2$. Из ОДЗ следует, что $t > 3/2$.<br>$t^2 + 4 = t(2t - 3)$<br>$t^2 + 4 = 2t^2 - 3t$<br>$t^2 - 3t - 4 = 0$</p><p>Решим квадратное уравнение. По теореме Виета, корни $t_1 = 4$ и $t_2 = -1$.</p><p>Проверим корни по условию $t > 3/2$:<br>1) $t_1 = 4$. $4 > 3/2$, корень подходит.<br>2) $t_2 = -1$. $-1 \ngtr 3/2$, корень не подходит.</p><p>Вернемся к замене с подходящим корнем:<br>$2^x = 4$<br>$2^x = 2^2$<br>$x = 2$</p><p>Проверим решение по ОДЗ ($x > log_2(1.5)$).<br>$2 > log_2(1.5)$, так как $log_2(4) > log_2(1.5)$. Решение подходит.</p><p>Ответ: $2$.</p>" ] #original: array:6 [ "id" => 1342178 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1141} "task" => array:2 [ "refs" => "104355" "type" => "task" ] "text" => "<p><strong>а)</strong> $log_2(9 - 2^x) = 3 - x$</p><p>Определим область допустимых значений (ОДЗ). Аргумент логарифма должен быть строго положительным:<br>$9 - 2^x > 0$<br>$9 > 2^x$<br>$log_2(9) > x$</p><p>Теперь решим уравнение. По определению логарифма ($log_a b = c \Leftrightarrow a^c = b$):<br>$9 - 2^x = 2^{3-x}$<br>$9 - 2^x = \frac{2^3}{2^x}$<br>$9 - 2^x = \frac{8}{2^x}$</p><p>Сделаем замену переменной. Пусть $t = 2^x$. Так как $2^x > 0$ для любого $x$, то $t > 0$.<br>$9 - t = \frac{8}{t}$<br>Умножим обе части на $t$ (так как $t \neq 0$):<br>$t(9 - t) = 8$<br>$9t - t^2 = 8$<br>$t^2 - 9t + 8 = 0$</p><p>Это квадратное уравнение. Решим его с помощью теоремы Виета или через дискриминант.<br>Корни уравнения: $t_1 = 1$, $t_2 = 8$. Оба корня положительны, поэтому оба подходят.</p><p>Вернемся к исходной переменной $x$:<br>1) $2^x = t_1 = 1 \Rightarrow 2^x = 2^0 \Rightarrow x_1 = 0$.<br>2) $2^x = t_2 = 8 \Rightarrow 2^x = 2^3 \Rightarrow x_2 = 3$.</p><p>Проверим, удовлетворяют ли найденные корни ОДЗ ($x < log_2(9)$).<br>$log_2(9) \approx 3.17$.<br>1) $x_1 = 0$. $0 < log_2(9)$, корень подходит.<br>2) $x_2 = 3$. $3 < log_2(9)$, так как $log_2(8) < log_2(9)$, корень подходит.</p><p>Ответ: $0; 3$.</p><p><strong>б)</strong> $log_2(25^x + 3 - 1) = 2 + log_2(5^x + 3 + 1)$</p><p>Упростим выражения в скобках:<br>$log_2(25^x + 2) = 2 + log_2(5^x + 4)$</p><p>ОДЗ:<br>1) $25^x + 2 > 0$. Это выражение всегда положительно, так как $25^x > 0$.<br>2) $5^x + 4 > 0$. Это выражение также всегда положительно, так как $5^x > 0$.<br>Следовательно, ОДЗ: $x \in \mathbb{R}$.</p><p>Преобразуем уравнение, используя свойства логарифмов:<br>$log_2(25^x + 2) - log_2(5^x + 4) = 2$<br>$log_2\left(\frac{25^x + 2}{5^x + 4}\right) = 2$</p><p>По определению логарифма:<br>$\frac{25^x + 2}{5^x + 4} = 2^2$<br>$\frac{25^x + 2}{5^x + 4} = 4$</p><p>Сделаем замену переменной. Пусть $t = 5^x$, тогда $25^x = (5^x)^2 = t^2$. Условие $t > 0$.<br>$\frac{t^2 + 2}{t + 4} = 4$<br>$t^2 + 2 = 4(t + 4)$<br>$t^2 + 2 = 4t + 16$<br>$t^2 - 4t - 14 = 0$</p><p>Решим квадратное уравнение через дискриминант:<br>$D = (-4)^2 - 4(1)(-14) = 16 + 56 = 72$<br>$t = \frac{4 \pm \sqrt{72}}{2} = \frac{4 \pm 6\sqrt{2}}{2} = 2 \pm 3\sqrt{2}$<br>Получаем два корня: $t_1 = 2 + 3\sqrt{2}$ и $t_2 = 2 - 3\sqrt{2}$.</p><p>Так как $t = 5^x$, то $t$ должно быть положительным. $t_1 = 2 + 3\sqrt{2} > 0$.<br>$t_2 = 2 - 3\sqrt{2} \approx 2 - 3 \cdot 1.41 = 2 - 4.23 < 0$. Этот корень не подходит.</p><p>Вернемся к замене: $5^x = 2 + 3\sqrt{2}$<br>$x = log_5(2 + 3\sqrt{2})$</p><p>Ответ: $log_5(2 + 3\sqrt{2})$.</p><p><strong>в)</strong> $log_4(2 \cdot 4^{x-2} - 1) = 2x - 4$</p><p>ОДЗ:<br>$2 \cdot 4^{x-2} - 1 > 0$<br>$2 \cdot \frac{4^x}{4^2} - 1 > 0$<br>$2 \cdot \frac{4^x}{16} - 1 > 0$<br>$\frac{4^x}{8} > 1$<br>$4^x > 8$<br>$(2^2)^x > 2^3 \Rightarrow 2^{2x} > 2^3 \Rightarrow 2x > 3 \Rightarrow x > \frac{3}{2}$</p><p>Решим уравнение, используя определение логарифма:<br>$2 \cdot 4^{x-2} - 1 = 4^{2x-4}$<br>Заметим, что $2x-4 = 2(x-2)$. Тогда $4^{2x-4} = 4^{2(x-2)} = (4^{x-2})^2$.</p><p>Сделаем замену $t = 4^{x-2}$. Уравнение примет вид:<br>$2t - 1 = t^2$<br>$t^2 - 2t + 1 = 0$<br>$(t - 1)^2 = 0$<br>$t = 1$</p><p>Вернемся к замене:<br>$4^{x-2} = 1$<br>$4^{x-2} = 4^0$<br>$x - 2 = 0$<br>$x = 2$</p><p>Проверим корень по ОДЗ ($x > 3/2$).<br>$2 > 1.5$, корень подходит.</p><p>Ответ: $2$.</p><p><strong>г)</strong> $log_2(4^x + 4) = log_2(2^x) + log_2(2^{x+1} - 3)$</p><p>ОДЗ:<br>1) $4^x + 4 > 0$ (верно для всех $x$).<br>2) $2^x > 0$ (верно для всех $x$).<br>3) $2^{x+1} - 3 > 0 \Rightarrow 2 \cdot 2^x > 3 \Rightarrow 2^x > \frac{3}{2} \Rightarrow x > log_2(\frac{3}{2})$.<br>Общее ОДЗ: $x > log_2(1.5)$.</p><p>Используем свойство суммы логарифмов $log_a b + log_a c = log_a(bc)$:<br>$log_2(4^x + 4) = log_2(2^x \cdot (2^{x+1} - 3))$</p><p>Так как основания логарифмов равны, приравниваем их аргументы:<br>$4^x + 4 = 2^x(2 \cdot 2^x - 3)$</p><p>Сделаем замену $t = 2^x$. Тогда $4^x = t^2$. Из ОДЗ следует, что $t > 3/2$.<br>$t^2 + 4 = t(2t - 3)$<br>$t^2 + 4 = 2t^2 - 3t$<br>$t^2 - 3t - 4 = 0$</p><p>Решим квадратное уравнение. По теореме Виета, корни $t_1 = 4$ и $t_2 = -1$.</p><p>Проверим корни по условию $t > 3/2$:<br>1) $t_1 = 4$. $4 > 3/2$, корень подходит.<br>2) $t_2 = -1$. $-1 \ngtr 3/2$, корень не подходит.</p><p>Вернемся к замене с подходящим корнем:<br>$2^x = 4$<br>$2^x = 2^2$<br>$x = 2$</p><p>Проверим решение по ОДЗ ($x > log_2(1.5)$).<br>$2 > log_2(1.5)$, так как $log_2(4) > log_2(1.5)$. Решение подходит.</p><p>Ответ: $2$.</p>" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "next" => array:2 [ "refs" => "104356" "type" => "task" ] "previous" => array:2 [ "refs" => "104354" "type" => "task" ] "book" => Illuminate\Database\Eloquent\Collection {#1105 #items: array:1 [ 0 => App\Models\Book {#1049} ] #escapeWhenCastingToString: false } "page" => Illuminate\Database\Eloquent\Collection {#1109 #items: array:1 [ 0 => App\Models\BookPage {#1107 #connection: "mysql" #table: "book_pages" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 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№524 (с. 245)
Условие. №524 (с. 245)
Решение 5. №524 (с. 245)
а) $log_2(9 - 2^x) = 3 - x$
Определим область допустимых значений (ОДЗ). Аргумент логарифма должен быть строго положительным:
$9 - 2^x > 0$
$9 > 2^x$
$log_2(9) > x$
Теперь решим уравнение. По определению логарифма ($log_a b = c \Leftrightarrow a^c = b$):
$9 - 2^x = 2^{3-x}$
$9 - 2^x = \frac{2^3}{2^x}$
$9 - 2^x = \frac{8}{2^x}$
Сделаем замену переменной. Пусть $t = 2^x$. Так как $2^x > 0$ для любого $x$, то $t > 0$.
$9 - t = \frac{8}{t}$
Умножим обе части на $t$ (так как $t \neq 0$):
$t(9 - t) = 8$
$9t - t^2 = 8$
$t^2 - 9t + 8 = 0$
Это квадратное уравнение. Решим его с помощью теоремы Виета или через дискриминант.
Корни уравнения: $t_1 = 1$, $t_2 = 8$. Оба корня положительны, поэтому оба подходят.
Вернемся к исходной переменной $x$:
1) $2^x = t_1 = 1 \Rightarrow 2^x = 2^0 \Rightarrow x_1 = 0$.
2) $2^x = t_2 = 8 \Rightarrow 2^x = 2^3 \Rightarrow x_2 = 3$.
Проверим, удовлетворяют ли найденные корни ОДЗ ($x < log_2(9)$).
$log_2(9) \approx 3.17$.
1) $x_1 = 0$. $0 < log_2(9)$, корень подходит.
2) $x_2 = 3$. $3 < log_2(9)$, так как $log_2(8) < log_2(9)$, корень подходит.
Ответ: $0; 3$.
б) $log_2(25^x + 3 - 1) = 2 + log_2(5^x + 3 + 1)$
Упростим выражения в скобках:
$log_2(25^x + 2) = 2 + log_2(5^x + 4)$
ОДЗ:
1) $25^x + 2 > 0$. Это выражение всегда положительно, так как $25^x > 0$.
2) $5^x + 4 > 0$. Это выражение также всегда положительно, так как $5^x > 0$.
Следовательно, ОДЗ: $x \in \mathbb{R}$.
Преобразуем уравнение, используя свойства логарифмов:
$log_2(25^x + 2) - log_2(5^x + 4) = 2$
$log_2\left(\frac{25^x + 2}{5^x + 4}\right) = 2$
По определению логарифма:
$\frac{25^x + 2}{5^x + 4} = 2^2$
$\frac{25^x + 2}{5^x + 4} = 4$
Сделаем замену переменной. Пусть $t = 5^x$, тогда $25^x = (5^x)^2 = t^2$. Условие $t > 0$.
$\frac{t^2 + 2}{t + 4} = 4$
$t^2 + 2 = 4(t + 4)$
$t^2 + 2 = 4t + 16$
$t^2 - 4t - 14 = 0$
Решим квадратное уравнение через дискриминант:
$D = (-4)^2 - 4(1)(-14) = 16 + 56 = 72$
$t = \frac{4 \pm \sqrt{72}}{2} = \frac{4 \pm 6\sqrt{2}}{2} = 2 \pm 3\sqrt{2}$
Получаем два корня: $t_1 = 2 + 3\sqrt{2}$ и $t_2 = 2 - 3\sqrt{2}$.
Так как $t = 5^x$, то $t$ должно быть положительным. $t_1 = 2 + 3\sqrt{2} > 0$.
$t_2 = 2 - 3\sqrt{2} \approx 2 - 3 \cdot 1.41 = 2 - 4.23 < 0$. Этот корень не подходит.
Вернемся к замене: $5^x = 2 + 3\sqrt{2}$
$x = log_5(2 + 3\sqrt{2})$
Ответ: $log_5(2 + 3\sqrt{2})$.
в) $log_4(2 \cdot 4^{x-2} - 1) = 2x - 4$
ОДЗ:
$2 \cdot 4^{x-2} - 1 > 0$
$2 \cdot \frac{4^x}{4^2} - 1 > 0$
$2 \cdot \frac{4^x}{16} - 1 > 0$
$\frac{4^x}{8} > 1$
$4^x > 8$
$(2^2)^x > 2^3 \Rightarrow 2^{2x} > 2^3 \Rightarrow 2x > 3 \Rightarrow x > \frac{3}{2}$
Решим уравнение, используя определение логарифма:
$2 \cdot 4^{x-2} - 1 = 4^{2x-4}$
Заметим, что $2x-4 = 2(x-2)$. Тогда $4^{2x-4} = 4^{2(x-2)} = (4^{x-2})^2$.
Сделаем замену $t = 4^{x-2}$. Уравнение примет вид:
$2t - 1 = t^2$
$t^2 - 2t + 1 = 0$
$(t - 1)^2 = 0$
$t = 1$
Вернемся к замене:
$4^{x-2} = 1$
$4^{x-2} = 4^0$
$x - 2 = 0$
$x = 2$
Проверим корень по ОДЗ ($x > 3/2$).
$2 > 1.5$, корень подходит.
Ответ: $2$.
г) $log_2(4^x + 4) = log_2(2^x) + log_2(2^{x+1} - 3)$
ОДЗ:
1) $4^x + 4 > 0$ (верно для всех $x$).
2) $2^x > 0$ (верно для всех $x$).
3) $2^{x+1} - 3 > 0 \Rightarrow 2 \cdot 2^x > 3 \Rightarrow 2^x > \frac{3}{2} \Rightarrow x > log_2(\frac{3}{2})$.
Общее ОДЗ: $x > log_2(1.5)$.
Используем свойство суммы логарифмов $log_a b + log_a c = log_a(bc)$:
$log_2(4^x + 4) = log_2(2^x \cdot (2^{x+1} - 3))$
Так как основания логарифмов равны, приравниваем их аргументы:
$4^x + 4 = 2^x(2 \cdot 2^x - 3)$
Сделаем замену $t = 2^x$. Тогда $4^x = t^2$. Из ОДЗ следует, что $t > 3/2$.
$t^2 + 4 = t(2t - 3)$
$t^2 + 4 = 2t^2 - 3t$
$t^2 - 3t - 4 = 0$
Решим квадратное уравнение. По теореме Виета, корни $t_1 = 4$ и $t_2 = -1$.
Проверим корни по условию $t > 3/2$:
1) $t_1 = 4$. $4 > 3/2$, корень подходит.
2) $t_2 = -1$. $-1 \ngtr 3/2$, корень не подходит.
Вернемся к замене с подходящим корнем:
$2^x = 4$
$2^x = 2^2$
$x = 2$
Проверим решение по ОДЗ ($x > log_2(1.5)$).
$2 > log_2(1.5)$, так как $log_2(4) > log_2(1.5)$. Решение подходит.
Ответ: $2$.
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