Помогите с решением log8 (2x-7)< 1. Created by nastёna1242001. algebra-ru

Помогите с решением log8 (2x-7)< 1...

nastёna1242001 @nastёna1242001

August 2022 2 2 Report

Помогите с решением log8 (2x-7)< 1

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paradiseva

Verified answer

$log_8(2x-7)\ \textless \ 1\\\\
 \left \{ {{log_8(2x-7)\ \textless \ 1} \atop {2x-7\ \textgreater \ 0}} \right. \\\\
 \left \{ {{log_8(2x-7)\ \textless \ log_8 8} \atop {2x\ \textgreater \ 7}} \right. \\\\
 \left \{ {{2x-7\ \textless \ 8} \atop {x\ \textgreater \ 3.5}} \right. \\\\
 \left \{ {{2x\ \textless \ 15} \atop {x\ \textgreater \ 3.5}} \right. \\\\
 \left \{ {{2x\ \textless \ 15} \atop {x\ \textgreater \ 3.5}} \right. \\\\
 \left \{ {{x\ \textless \ 7.5} \atop {x\ \textgreater \ 3.5}} \right. \\
$

Ответ: $x \in (3.5;7.5)$

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nastёna1242001 Эмммм

nastёna1242001 Спасибо огромное

zinaidazina Log₈ (2x-7) < 1

ОДЗ: 2х - 7 > 0
x > 3,5

1) log₈ (2x-7) < 1
log₈ (2x-7) < log₈8
2x - 7 < 8
2x < 8 + 7
2x < 15
x < 7,5

2) {x > 3,5
{x < 7,5
║
∨
3,5 < x < 7,5

x ∈ ] 3,5; 7,5[