The other answers are valid, but require you to spot that 3125=5^{5} which is not always practical.
A quicker alternative (that will work equally well for less convenient numbers) is to use a logarithm: if b^{x}=a then log_{b}a=x, so
log_{25}3125 = x = 2.5
or, if you can only calculate natural logs for whatever reason,
x = log_{25}3125 = log(3125)/log(25) = ~3.495/~1.398 = 2.5
25^x=3125
we will take the L.C.M of 3125 i.e. 5*5*5*5*5=5^5
(5^2)^X= 5^5
now (a^m)^n=a^mn
we will put the problem in the given formula then
5^2x=5^5
now, a^m=a^n =>m=n
according to formula, 2x=5
= x=5/2
25^{x} = 3,125 ----> This is an exponential equation (How can you get the value of x?)
5^{2(x)} = 5^{5} -----> For this equation, the base is 5 and the exponents for 25 and 3,125 are 2 and 5 respectively.
2x = 5 -----> remove the base 5, and the value of x is:
x = 5/2
25^x = ((5)^2)^x = 5^2x
3125 = ((25)^2) * 5 = 5^5
equating two above equations we get
2x= 5
x = 5/2 = 2.5