How do you solve $-2log_5(7x)=2$ ?

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Answer 1 · 2016-07-27T10:26:20

$x=1/35$

You can divide all terms by 2 and have:

$log_5(7x)=-1$

you can note that

$-1=log_5(1/5)$

so you can substitute:

$log_5(7x)=log_5(1/5)$

that's equivalent to:

$7x=1/5 and x>0$

$x=1/35$

How do you solve -2log_5(7x)=2-2log_5(7x)=2?