Help for how to solve for x using logs with different bases

log(base5)3^x = log(base25)9^(1-2x)

Thank you in advance ! (:

log(base5)3^x = log(base25)9^(1-2x)

x log(base5)3= (1-2x) log(base25)9

log(base25)9= log(base5^2)3^2=log(base5)3

x=1-2x
x=1/3


attention: in general case

log(base a)(b)= log(base c)(b) / log(base c)(a)

xlog(5)3 . . . by the properties of logarithms
(1-2x)log(25)3² . . . ditto
3(1-2x)log(25)3 . . . ditto

xlog(5)3 = 3(1-2x)log(25)3

x / 3(1-2x) = log(25)3/log(5)3 . . .change of base and rearranging

x / 3(1-2x) = (log3/log25)/log3/log5) = log5/log25 = log5/2log5 = 1/2

x / 3(1-2x) = 1/2