You should use the logarithms as you use the squaring when you need to remove a square root such that:
Notice that the logarithms have the property to bring down the exponent, as it was proved above.
The way of solving the exponential equation depends on the right term b.
If b is a multiple of a, then the equation could be solved using the one to one property of exponential function.
Let's see how:
2^2x = 8
But 8 is a multiple of 2:
8 = 2^3
2^2x = 2^3
Since the bases are matching, we'll apply one to one property:
2x = 3
x = 3/2
If b is not a multiple of a, then the equation could be solved using logarithms.
2^3x = 3
Since we cannot create matching bases, we'll take logarithms both sides:
lg 2^3x = lg 3
We'll apply product property of logarithms:
3x*lg2 = lg3
3x = lg3/lg2
x = lg3/3lg2
x = lg3/lg8