2ish methods -
#1 - get common basis
"Solve 2x = 32" ... convert the 32 to 25, set the exponents equal, and solve for "x = 5"
#2 - use logarithms
Solve 2x = 30
logb(mn) = n · logb(m)
When I take the log of both sides of an equation, I can use any log I like (base-10 log, base-2 log, natural log, etc), but some are sometimes more useful than others. Since the base in the equation "2x = 30" is "2", I might try using a base-2 log:
log2(2x) = log2(30)
Any log of the log's base returns a value of 1, so log2(2) = 1. Then:
x · log2(2) = log2(30)
x(1) = log2(30)
x = log2(30)
#3 - use natural log
2x = 30
ln(2x) = ln(30)
x · ln(2) = ln(30)
x = ln(30)​/ln(2)