Answer:
Step-by-step explanation:
There is a possible issue in your question...
As written it can be interpreted in two ways
like this #1 :
[tex]5^{2x} = 3(2^x)[/tex]
or like this #2 :
[tex]5^2x = 3(2^x)[/tex]
VERSION #1
divide by the 2^x
[tex]\frac{5^{2x} }{2^x} = 3[/tex]
to be able to get to the exponent you have to take "logs"
... look up the rules for logs ... log(ab) = log (a)+log(b) , log(a/b) = log(a)-log(b) etc.
if you take the logs of both sides the using the quotient rule and the exponent rule ....result the result is...
[tex]5^{2x}=3\left(2^x\right)\quad :\quad x=\frac{\ln \left(3\right)}{2\ln \left(5\right)-\ln \left(2\right)}\quad \left(\mathrm{Decimal}:\quad x=0.43496\dots \right)[/tex]
VERSION #2
[tex]5^2x\:=\:3\left(2^x\right)[/tex]
25 x = 3(2^x)
[tex]\frac{25}{3} = \frac{2^{x} }{x}[/tex]
This gets really nasty, and I assume that it is not the original problem..
[tex]x=-\frac{\text{W}_{-1}\left(-\frac{3\ln \left(2\right)}{25}\right)}{\ln \left(2\right)},\:x=-\frac{\text{W}_0\left(-\frac{3\ln \left(2\right)}{25}\right)}{\ln \left(2\right)}\quad \mathrm{ }:\quad x=5.52482 ,\:x=0.13144[/tex]
