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: 16

Answer

$x=\dfrac{\log15}{\log9}\approx1.232487$

Work Step by Step

$3^{2x-1}=5$ Using the product of powers rule, we can write the left side of the equation like this: $(3^{2x})(3^{-1})=5$ We know that $(a^{b})^{c}=a^{bc}$ and $a^{-n}=\dfrac{1}{a^{n}}$. Applying this to the left side, it becomes: $(3^{2})^{x}(\dfrac{1}{3})=5$ $(9^{x})(\dfrac{1}{3})=5$ Take the fraction's denominator to multiply the right side of the equation: $9^{x}=15$ Apply $\log$ to both sides: $\log9^{x}=\log5$ The exponent $x$ can be taken down to multiply in front of the $\log$: $x\log9=\log15$ Solve for $x$: $x=\dfrac{\log15}{\log9}\approx1.232487$
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