Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 368: 28

Answer

$x=\dfrac{\log45}{\log3}-1\approx2.464974$

Work Step by Step

$2(5+3^{x+1})=100$ First, let's solve for $3^{x+1}$: $5+3^{x+1}=\dfrac{100}{2}$ $5+3^{x+1}=50$ $3^{x+1}=50-5$ $3^{x+1}=45$ Apply $\log$ to both sides of the equation: $\log3^{x+1}=\log45$ The exponent $x+1$ can be taken down to multiply in front of its respective $\log$: $(x+1)\log3=\log45$ Solve for $x$: $x+1=\dfrac{\log45}{\log3}$ $x=\dfrac{\log45}{\log3}-1\approx2.464974$
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