College Algebra 7th Edition

Published by Brooks Cole

Chapter 4, Exponential and Logarithmic Functions - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises: 49

Answer

$x=5$

Work Step by Step

$\log x+\log(x-1)=\log(4x)$ $\log[x(x-1)]=\log(4x)$ $x(x-1)=4x$ $x^{2}-x=4x$ $x^{2}-x-4x=0$ $x^{2}-5x=0$ $x(x-5)=0$ $x=0$ or $x-5=0$ $x=0$ or $x=5$ But $x=0$ is not possible in the original equation, because we can't take a log of zero (undefined).

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