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: 34

Answer

$x=\dfrac{1}{1+\log6}\approx0.562382$

Work Step by Step

$10^{1-x}=6^{x}$ Apply $\log$ to both sides of the equation: $\log10^{1-x}=\log6^{x}$ The exponents $1-x$ and $x$ can be taken down to multiply in front of their respective logarithms: $(1-x)\log10=x\log6$ Since $\log10=1$, the equation becomes: $1-x=x\log6$ Solve for $x$: $x\log6+x=1$ $x(\log6+1)=1$ $x=\dfrac{1}{1+\log6}\approx0.562382$
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