Solving Exponential Equations with Logarithms

If this equation had asked me to "Solve 2^x = 32", then finding the solution would have been easy, because I could have converted the 32 to 2⁵, set the exponents equal, and solved for "x = 5". But, unlike 32, 30 is not a power of 2 so I can't set powers equal to each other. I need some other method of getting at the x, because I can't solve with the equation with the variable floating up there above the 2; I need it back down on the ground where it belongs, where I can get at it. And I'll have to use logarithms to bring that variable down.

When dealing with equations, I can do whatever I like to the equation, as long as I do the same thing to both sides. And, to solve an equation, I have to get the variable by itself on one side of the "equals" sign; to isolate the variable, I have to "undo" whatever has been done to the variable.

In this case, the variable x has been put in the exponent. The backwards (technically, the "inverse") of exponentials are logarithms, so I'll need to undo the exponent by taking the log of both sides of the equation. This is useful to me because of the log rule that says that exponents inside a log can be turned into multipliers in front of the log:

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 "2^x = 30" is "2", I might try using a base-2 log:

Any log of the log's base returns a value of 1, so log₂(2) = 1. Then:

x = log₂(30)

2^x = 30

ln(2^x) = ln(30)

x · ln(2) = ln(30)

Since 212 is not a power of 5, then I will have to use logs to solve this equation. I could take base-5 log of each side, solve, and then apply the change-of-base formula, but I think I'd rather just use the natural log in the first place:

Since 52 is not a power of 10, I will have to use logs to solve this. In this particular instance, since the base is 10 and since base-10 logs can be done on the calculator, I will use the common log instead of the natural log to solve this particular equation:

Before I can start looking at the exponential, I first have to get rid of the 3, so I'll divide that off to get:

Since is not a power of 2, I will have to use logs. I will use the natural log in this case:

2^x+4 = (2^x)(2⁴) = (2^x)(16)