Question

What is the true solution to the equation below?

Answer 1

Yeeee

assuming your equaiton is
$2ln(e^{ln(2x)})-ln(e^{ln(10x)})=ln(30)$


remember some nice log rules
$log_a(b)=c$ translates to $a^c=b$
and
$a^{log_a(b)}=b$
and
$xlog_c(b)=log_c(b^x)$
and
$ln(x)=log_e(x)$
and
$log(a)-log(b)=log(\frac{a}{b})$
and
if $log(a)=log(b)$ then a=b

so

we can simplify a bit of stuff here

the $e^{ln(2x)} \space\ and \space\ the \space\ e^{ln(10x)}$ can be simplified to $2x \space\ and \space\ 10x$

so we gots now

$2ln(2x)-ln(10x)=ln(30)$
$ln((2x)^2)-ln(10x)=ln(30)$
$ln(4x^2)-ln(10x)=ln(30)$
$ln(\frac{4x^2}{10x})=ln(30)$
same base so
$\frac{4x^2}{10x}=30$
$\frac{2x}{5}=30$
times both sides by 5
$2x=150$
divide both sides by 2
$x=75$
answer is x=75