How do you solve $2log(2x) = 1 + loga$ ?

Question

3659 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-15T01:56:03

$x=sqrt(10a)/2$ (Assuming $log = log_10$ )

$2log_10(2x) = 1+log_10 a$

$2log_10 2x - log_10 a =1$

$2log_10 2x - 2log_10 a^(1/2) =1$

$2log_10((2x)/sqrt(a)) =1$

$log_10((2x)/sqrt(a)) =1/2$

$(2x)/sqrt(a) = 10^(1/2)$

$2x=sqrt(10a)$

$x=sqrt(10a)/2$

How do you solve 2log(2x) = 1 + loga2log(2x) = 1 + loga?