How do you solve $- 2 {log}_{5} (7 x) = 2$ $-2log_5(7x)=2$ ?

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Answer 1 · 2016-07-27T10:26:20

$x = \frac{1}{35}$ $x=1/35$

You can divide all terms by 2 and have:

${log}_{5} (7 x) = - 1$ $log_5(7x)=-1$

you can note that

$- 1 = {log}_{5} (\frac{1}{5})$ $-1=log_5(1/5)$

so you can substitute:

${log}_{5} (7 x) = {log}_{5} (\frac{1}{5})$ $log_5(7x)=log_5(1/5)$

that's equivalent to:

$7 x = \frac{1}{5} and x > 0$ $7x=1/5 and x>0$

$x = \frac{1}{35}$ $x=1/35$

How do you solve −2log5(7x)=2-2log_5(7x)=2?