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Solving Equations

There are several types of equations that arise with logarithms that you should be able to solve. The list below gives examples of the various types. There are two formulas you should be aware of when working on this problem, both of them are true because the logarithm and inverse functions are inverses of each other.

blogbx = xandlogbbx = x

I wrote the formulas using a generic base, b, to emphasise the point that they work for any value. I'll rewrite them again using the default base 10 just because that may be easier to read.

10log(x) = xandlog(10x) = x

Example log3.1(x) = log3.1(2x - 7)
Solution

Don't be thrown by the strange base. The rule here is a simple one: If the bases are the same then the values have to be as well. This turns our equation into

x = 2x - 7
-x = -7
x = 7

Example 4.235x + 2 = 4.233x
Solution

The situation here is identical to the last problem only with exponents instead of logarithms. If the bases's are the same then the exponents have to be so our equation becomes.

5x + 2 = 3x
2x + 2 = 0
2x = -2
x = -1

Example log3(x) = 4
Solution

Remember that the logarithm part means, "4 is the power I have to raise 3 to to get x". Translating this into an equation gives us

34 = x
x = 81

Example log(x2) = 10
Solution

First, I'm going to move the exponent off of the two using the first of our logarithm rules

log(x2) = 10
2 log(x) = 10
log(x) = 5

Now I'm going to apply the definition of a logarithm which says "5 is the power I have to raise 10 to in order to get x".

x = 105 = 100,000

Example 2log(x) = log(3) + log(2x - 3)
Solution

Our goal in most of these problems is to get rid of the logarithms, leaving us with a regular polynomial or linear equation. In this case, I'm going to pull the 2 on the left hand side back into the logarithm

2log(x) = log(x2)

Then I'm going to use our arithmetic rule to combine the terms on the right.

log(3) + log(2x - 3) = log(3(2x - 3)) = log(6x - 9)

Substituting these back into our original equation turns it into

log(x2) = log(6x - 9)

But now this is the same type as the first problem on this page. Because the logarithms's bases are the same we can set the expressions equal to each other which gives us a polynomial equation.

x2 = 6x - 9
x2 - 6x + 9 = 0
(x - 3)2 = 0
x - 3 = 0
x = 3