# Algebra Tricks: How to Solve Logarithm Equations

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 log_{10} (log base 10) or ln (natural log base e) then the following two identities are also useful:

- e^ln(x) = x
- 10^log
_{10}(x) = x

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

- ln(x)/ln(m) = log
_{m}(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*log_{10}(x) = 2 + log_{10}(x-1) + log_{10}(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**.

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