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: 32

Answer

$x=\dfrac{\log\dfrac{100}{3}}{\log125}\approx0.726249$

Work Step by Step

$125^{x}+5^{3x+1}=200$ Use the product of powers rule to rewrite the left side of the equation like this: $125^{x}+(5^{3x})(5)=200$ Now, use the power of a power rule to rewrite $5^{3x}$ like this: $125^{x}+(5)(5^{3})^{x}=200$ $125^{x}+(5)(125^{x})=200$ Solve for $125^{x}$ $(6)125^{x}=200$ $125^{x}=\dfrac{200}{6}$ $125^{x}=\dfrac{100}{3}$ Apply $\log$ to both sides of the equation: $\log125^{x}=\log\dfrac{100}{3}$ The exponent $x$ can be taken down t multiply in front of the $\log$: $x\log125=\log\dfrac{100}{3}$ Solve for $x$: $x=\dfrac{\log\dfrac{100}{3}}{\log125}\approx0.726249$
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