yeeee
assuming your equaiton is
[tex]2ln(e^{ln(2x)})-ln(e^{ln(10x)})=ln(30)[/tex]
remember some nice log rules
[tex]log_a(b)=c[/tex] translates to [tex]a^c=b[/tex]
and
[tex]a^{log_a(b)}=b[/tex]
and
[tex]xlog_c(b)=log_c(b^x)[/tex]
and
[tex]ln(x)=log_e(x)[/tex]
and
[tex]log(a)-log(b)=log(\frac{a}{b})[/tex]
and
if [tex]log(a)=log(b)[/tex] then a=b
so
we can simplify a bit of stuff here
the [tex]e^{ln(2x)} \space\ and \space\ the \space\ e^{ln(10x)}[/tex] can be simplified to [tex]2x \space\ and \space\ 10x[/tex]
so we gots now
[tex]2ln(2x)-ln(10x)=ln(30)[/tex]
[tex]ln((2x)^2)-ln(10x)=ln(30)[/tex]
[tex]ln(4x^2)-ln(10x)=ln(30)[/tex]
[tex]ln(\frac{4x^2}{10x})=ln(30)[/tex]
same base so
[tex]\frac{4x^2}{10x}=30[/tex]
[tex]\frac{2x}{5}=30[/tex]
times both sides by 5
[tex]2x=150[/tex]
divide both sides by 2
[tex]x=75[/tex]
answer is x=75
