College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check: 61

Answer

$x\approx -1.15$

Work Step by Step

Take the natural log of both sides to obtain: $\ln{(4^{1-x})} = \ln{(3^{2x+5})}$ Use the rule $\log{(a^n)} = n\cdot\log{a}$ to obtain: $(1-x)\ln{4} = (2x+5)\ln{3}$ Distribute $ln{4}$ on the left and $\ln{3}$ on the right to obtain: $\ln{4} - x\ln{4} = 2x\ln{3} + 5\ln{3}$ Subtract $5\ln3$ on both sides of the equation to obtain: $\ln4 - x\ln4 - 5\ln3 = 2x\ln3+5\ln3 - 5\ln3 \\\ln4 - x\ln4 - 5\ln3 = 2x\ln3$ Add $x\ln4$ to both sides of the equation to obtain: $\ln4 - x\ln4 - 5\ln3 +x\ln4= 2x\ln3+x\ln4 \\\ln4 - 5\ln3 = 2x\ln3+x\ln4$ Factor out $x$ on the right side of the equation to obtain: $\ln4 -5\ln3 = x(2\ln3+\ln4)$ Divide $\ln3+\ln4$ to both sides of the equation to obtain: $\dfrac{\ln4-5\ln3}{2\ln3 + \ln4}=\dfrac{x(2\ln3+\ln4)}{2\ln3+\ln4} \\\dfrac{\ln4-5\ln3}{2\ln3 + \ln4}=x$ Use a calculator to obtain: $x\approx -1.15$
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