how to solve 3^x+2 = 5^2x?

Have to assume that you mean :-

3^(x + 2) = 5^(2x)

( x + 2 ) log 3 = (2x) log 5

2 log 3 = (2x) log 5 - x log 3

2 log 3 = ( 2 log 5 - log 3 ) x

x = 2 log 3 / ( 2 log 5 - log 3 )

x = 1.036

3^(x + 2) = 5^(2x)
log[3^(x + 2)] = log[5^(2x)]
(x + 2)log(3) = (2x)log(5)
(x)log(3) + 2log(3) = (2x)log(5)
(x)log(3) - (2x)log(5) = -2log(3)
x[log(3) - 2log(5)] = -2log(3)
x = [-2log(3)]/[log(3) - 2log(5)]

3^(x+2) = 5^2x
taking log of both sides
(x+2)log3 =(2x)log5
or 2log3=2xlog5 -xlog3
or 2log3=x(2log5 -log3)
or log3^2 =x(log5^2 -log3)
or log9 =xlog(25/3)
or 0.9542 =0.9218 x
so x =0.9542/0.9218
giving x= 1.04

3^(x + 2) = 5^2x
3^x * 3^2 = 25^x
9 = 25^x/3^x
9 = (25/3)^x
x = log(25/3)_9 = 1.036297866

next time please use brackets like this:
3^(x + 2) = 5^2x
not like this
3^x + 2 = 5^2x
people may assume that 2 is not part of the power
s - t - u - p - i - d

3^(x+2) = 5^2x
3^2 * 3^x =(25)^x
(25/3)^x = 9
x = log 9/log(25/)=
2log 3/(2log5-log3)=
2log 3/(2log 5- log 3)=1.036297866
(approx)

i suck too and you shouldent be cheating online