Question

log8^x+log4^x+log2^x=1 find x​

Answer 1

Step-by-step explanation:

Given,

$log( {8}^{x} ) + log( {4}^{x} ) + log( {2}^{x} ) = 1$

$= > log( {2}^{3x} ) + log( {2}^{2x} ) + log( {2}^{x} ) = 1$

$= > 3x \: log(2) + 2x \: log(2) + x \: log(2) = 1$

$= > 6x \: log(2) = 1$

$= > x = \frac{1}{ log( {2}^{6} ) }$

$= > x = log_{32}(e)$