Question

log
$log {8}^{x} + log {4}^{x} + log {2}^{x} = 11$
​

Answer 1

$\boxed{\underline{\textsf{Formulas :}}}$

$1.\:\log\:a^{b}=b\log\:a$

$2.\: \log\:a+\log\:b+\log\:c = \log\:(abc)$

$\boxed{\underline{\textsf{Solution :}}}$

$\textsf{Now,}\:\log8^{x}+\log4^{x}+\log2^{x}=11$

$\to x\log8+x\log4+x\log2=11$

$\to x (\log8+\log4+\log2)=11$

$\to x\log(8*4*2)=11$

$\to x\log64=11$

$\to x = \frac{11}{\log64}$

$\to x = \frac{11}{1.8},\:\textsf{since}\:\log64\approx1.8$

$\to \boxed{\bold{x \approx 6.1}}$

$\textsf{which is the required solution.}$

$\underline{\textsf{Finding log64}}$

$\log64$

$=\log2^{6}$

$=6\log2$

$\approx 6*0.30,\:since\:\log2 \approx 0.30$

$\bold{\approx 1.8}$