Question

log2 x+log4 x+log8 x=11

Answer 1

ОДЗ : x > 0

$log_{2}x+log_{4}x+log_{8}x=11 \\\\log_{2}x+log_{2}x^{\frac{1}{2}}+log_{2}x^{\frac{1}{3}} =11\\\\log_{2}x+\frac{1}{2} log_{2}x+\frac{1}{3}log_{2}x=11\\\\\frac{11}{6}log_{2}x=11\\\\log_{2} x=6\\\\x=2^{6} \\\\\boxed{x=64}$

Answer 2

Ответ:

$64$

Объяснение:

ОДЗ:

$x>0;$

Решение:

$log_{2}x+log_{4}x+log_{8}x=11;$

$log_{2^{1}}x+log_{2^{2}}x+log_{2^{3}}x=11;$

$\dfrac{1}{1}log_{2}x+\dfrac{1}{2}log_{2}x+\dfrac{1}{3}log_{2}x=11;$

$(1+\dfrac{1}{2}+\dfrac{1}{3}) \cdot log_{2}x=11;$

$(\dfrac{6}{6}+\dfrac{3}{6}+\dfrac{2}{6}) \cdot log_{2}x=11;$

$\dfrac{11}{6} \cdot log_{2}x=11;$

$log_{2}x=11:\dfrac{11}{6};$

$log_{2}x=11 \cdot \dfrac{6}{11};$

$log_{2}x=6;$

$x=2^{6};$

$x=64;$

Проверка:

$log_{2}64+log_{4}64+log_{8}64=log_{2}2^{6}+log_{4}4^{3}+log_{8}8^{2}=6+3+2=11;$