## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 4 - Review - Exercises: 61

#### Answer

$x=-\dfrac{5\log3-\log4}{\log4+2\log3}\approx-1.15$

#### Work Step by Step

$4^{1-x}=3^{2x+5}$ Apply $\log$ to both sides of the equation: $\log4^{1-x}=\log3^{2x+5}$ Take $1-x$ and $2x+5$ down to multiply in front of their respective logarithms: $(1-x)\log4=(2x+5)\log3$ Evaluate the products: $\log4-x\log4=2x\log3+5\log3$ Take $2x\log3$ to the left side and $\log4$ to the right side: $-x\log4-2x\log3=5\log3-\log4$ Take out common factor $x$ from the left side: $x(-\log4-2\log3)=5\log3-\log4$ Solve for $x$: $x=-\dfrac{5\log3-\log4}{\log4+2\log3}\approx-1.15$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.