# How to Find Missing Exponents ••• glasses and mathematics image by Alexander Potapov from Fotolia.com

Solving for a missing exponent can be as simple as solving 4=2^x, or as complex as finding how much time must pass before an investment is doubled in value. (Note that the caret refers to exponentiation.) In the first example, the strategy is to rewrite the equation so both sides have the same base. The latter example may take the form principal_(1.03)^years for the amount in an account after earning 3 percent annually for a certain number of years. Then the equation to determine the time to doubling is principal_(1.03)^years=2*principal, or (1.03)^years=2. One then needs to solve for the exponent "years (Note that asterisks denote multiplication.)

## Basic Problems

Move the coefficients over to one side of the equation. For example, suppose you need to solve 350,000=3.5*10^x. Then divide both sides by 3.5 to get 100,000=10^x.

Rewrite each side of the equation so the bases match. Continuing with the example above, both sides can be written with a base of 10. 10^6 = 10^x. A harder example is 25^2=5^x. The 25 can be rewritten as 5^2. Note that (5^2)^2=5^(2*2)=5^4.

Equate the exponents. For example, 10^6=10^x means x must be 6.

## Using Logarithms

Take the logarithm of both sides instead of making the bases match. Otherwise, you may have to use a complex logarithm formula to make the bases match. For example, 3=4^(x+2) would need to be changed into 4^(log 3/log 4)=4^(x+2). The general formula for making bases equal is: base2=base1^(log base2 / log base1). Or you could just take the log of both sides: ln 3=ln [4^(x+2)]. The base of the logarithm function you use does not matter. The natural log (ln) and the base-10 log are equally fine, as long as your calculator can calculate the one you pick.

Bring the exponents down in front of the logarithms. The property being used here is log (a^b)=b_log a. This property can intuitively be seen to be true if you now that log ab=log a + log b. This is because, for example, log (2^5)=log(2_2_2_2_2)=log2+log2+log2+log2+log2=5log2. So for the doubling problem stated in the introduction, log (1.03)^years=log 2 becomes years_log (1.03)=log 2.

Solve for the unknown like any algebraic equation. Years=log 2 / log(1.03). So to double an account paying an annual rate of 3 percent, one must wait 23.45 years.