How do you solve $log_4x=log_8(4x)$ ?

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Answer 1 · 2016-08-23T09:24:24

$x = 2^4$

Using

$log_a b= log_e a/(log_e b)$

$log_4x = (log_e x)/(log_e 4) = log_8(4x) = (log_e (4x))/(log_e 8)$

but $log_e 4 = 2 log_e 2$ and $log_e 8 = 3 log_e 2$ so

$log_e x/(2log_e 2) = (log_e x + 2log_e 2)/(3 log_e 2)$

$3log_e x = 2log_e x+4 log_e 2$

and finally

$log_e x = log_e 2^4$

so

$x = 2^4$

How do you solve log_4x=log_8(4x)log_4x=log_8(4x)?