Example

Example 10.39: Solving log3x+log3(x8)=2\log_3 x + \log_3(x-8) = 2

To solve the logarithmic equation log3x+log3(x8)=2\log_3 x + \log_3(x-8) = 2, first use the Product Property of Logarithms to condense the left side into a single logarithm: log3(x(x8))=2\log_3(x(x-8)) = 2. Next, convert this to its equivalent exponential form: 32=x(x8)3^2 = x(x-8). Simplifying both sides gives 9=x28x9 = x^2 - 8x. Subtract 9 from each side to set the quadratic equation to zero: 0=x28x90 = x^2 - 8x - 9. Factoring the right side yields 0=(x9)(x+1)0 = (x-9)(x+1). Using the Zero-Product Property, the potential solutions are x=9x = 9 and x=1x = -1. However, substituting x=1x = -1 into the original equation results in evaluating the logarithm of a negative number, which is undefined. Therefore, x=1x = -1 is an extraneous solution and is eliminated. The only valid solution is x=9x = 9. Checking x=9x = 9 by substitution gives log39+log31=2+0=2\log_3 9 + \log_3 1 = 2 + 0 = 2, confirming the result.

0

1

Updated 2026-05-26

Contributors are:

Who are from:

Tags

OpenStax

Intermediate Algebra @ OpenStax

Ch.10 Exponential and Logarithmic Functions - Intermediate Algebra @ OpenStax

Algebra

Related