How do you solve these logarithms?

Question

$2 log_(2)(1-2x)=12$
$log(7x)+4=5$

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Answer 1 · 2016-11-26T22:44:12

$x=(1pm2^6)/2$
$x=e/7$

Before the $log$ application we had:

$(1-2x)^2=2^12$
$7xe^4=e^5$

so

$1-2x=pm2^6->x=(1pm2^6)/2$

and

$x=e/7$