# Algebra Tricks: How to Solve Logarithm Equations

Solve logarithm equations with algebra.

To solve algebraic equations involving logarithms, it is essential to know the basic logarithm identities:

• log(a*b) = log(a) + log(b)
• log(a/b) = log(a) - log(b)
• log(a^b) = b*log(a)
• log(1) = 0

where "log" denotes the logarithm in any base. If you are dealing with log10 (log base 10) or ln (natural log base e) then the following two identities are also useful:

• e^ln(x) = x
• 10^log10(x) = x

If you need to switch between bases, the logarithm base change identity is

• ln(x)/ln(m) = logm(x)

## Example Problem 1

As long as you know the important logarithm identities you can solve any log equation or determine if it has no solution. For example, use the expressions above to simplify ln(5x+1) - ln(x-5) = ln(2).

The first step is to simplify the right hand side:

ln[(5x+1)/(x-5)] = ln(2)

Now use the identity relation e^ln(x) = x:

e^ln[(5x+1)/(x-5)] = e^ln(2)

(5x+1)/(x-5) = 2.

This rational function can be solved using simple algebra

(5x+1)/(x-5) = 2

5x + 1 = 2x - 10

3x = -11

x = -11/3

The final step is to plug this answer back into the original equation to make sure the solution does not yield an undefined quantity. In this example, ln(-11/3 - 5) is undefined, so the original problem has no solution.

## Example Problem 2

Solve 2*log10(x) = 2 + log10(x-1) + log10(x-1.5).

First, put all the terms involving x on the left side and the constant term on the right:

2*log10(x) - [log10(x-1) + log10(x-1.5)] = 2

Now simplify the left side using the basic log rules and the quadratic formula:

log(x^2) - log[(x-1)(x-1.5)] = 2

log(x^2/[(x-1)(x-1.5)]) = 2

10^log(x^2/[(x-1)(x-1.5)]) = 10^2

x^2/[(x-1)(x-1.5)] = 100

x^2 = 100(x-1)(x-1.5)

x^2 = 100x^2 - 250x + 150

0 = 99x^2 - 250x + 150

x = 1.5438, 0.9814

Only the first solution works; the second yields an undefined quantity. Thus, the answer is x = 1.5438.