## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 12 - Exponential Functions and Logarithmic Functions - 12.6 Solving Exponential Equations and Logarithmic Equations - 12.6 Exercise Set - Page 825: 48

#### Answer

$x=1$

#### Work Step by Step

Using the laws of logarithms, the given equation, $\log(x+9)+\log x=1 ,$ is equivalent to \begin{array}{l}\require{cancel} \log[(x+9)\cdot x]=1 \\\\ \log(x^2+9x)=1 .\end{array} In exponential form, the logarithmic equation, $\log(x^2+9x)=1 ,$ is equivalent to \begin{array}{l}\require{cancel} \log_{10}(x^2+9x)=1 \\\\ x^2+9x=10^1 .\end{array} Solving the equation, $x^2+9x=10^1 ,$ results in \begin{array}{l}\require{cancel} x^2+9x=10 \\\\ x^2+9x-10=0 \\\\ (x+10)(x-1)=0 \\\\ x=\{-10,1\} .\end{array} Upon checking only $x=1$ satisfies the original equation.

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