It varies from problem to problem, but the two common ways to solve exponential equations are either by "equating the bases" or using logarithms to break them down. For example, if you have something like 5ˣ - 1 = 6, you only have one exponent involved, so there aren't any bases to equate.
5ˣ - 1 = 6 ... add 1 to both sides
5ˣ = 7 ... you have one term on either side, so take the log of both sides
log 5ˣ = log 7 ... by exponent properties, you can now bring that x out front
x log 5 = log 7 ...solve for x, divide both sides by log 5
x = (log 7)/(log 5)
That's a really simple example, but it applies the same to problems like 5ˣ⁺³ - 1 = 6, too. Same exact process. Get the exponent alone, take the log of both sides, solve for x.
For equating the bases, if you have something like
2ˣ = 4ˣ⁺²
you can recognize that 4 and 2 are multiples of one another. In fact, 4 = 2². So in this case, you play around with squares to get something where the bases of the exponents (i.e. the number being squared, or cubed, or set to the xth power) match.
2ˣ = 4ˣ⁺² ... but 4 = 2^2, so write it in this form instead
2ˣ = (2²)ˣ⁺² ... by exponent rules, you multiply the exponents on the righthand side, so distribute 2 to x + 2
2ˣ = 2²ˣ⁺⁴
And problems like this are really nice, if you can equate the bases using square roots and cube roots, because what you can do now is totally get rid of the 2's on either side and just drop it down to x = 2x + 4, then solve for x. The math behind this is:
2ˣ = 2²ˣ⁺⁴ ... single exponent on either side, so you can take the log of both sides
log 2ˣ = log 2²ˣ⁺⁴ ... bring the exponents out front
x log 2 = (2x + 4) log 2 ... divide both sides by log 2
x = 2x + 4
So once you know that property, you can skip a little work. Solving an exponential equation for x almost always involves a logarithm, and you should look into properties of exponents. They'll help you manipulate the problem by writing stuff in different forms so you can find the answer.
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