Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.6 Logarithmic and Exponential Equations - 4.6 Assess Your Understanding - Page 337: 52


$\text{Exact: } \dfrac{ \ln \left(\dfrac{4}{3} \right)}{\ln 5 +\ln \left(\dfrac{4}{3} \right)}$ $\text{Approximately: } 0.152$

Work Step by Step

Take the natural logarithm of both sides: $\ln \left(\dfrac{4}{3} \right)^{1-x} = \ln 5^{x}$ Since $\ln M^r = r \ln M$, then: $\therefore\ln \left(\dfrac{4}{3} \right)^{1-x} = (1-x) \ln \left(\dfrac{4}{3} \right) \hspace{20pt} \text{and} \hspace{20pt} \ln 5^{x} = x \ln 5$ Thus, the equation above is equivalent to : $(1-x) \ln \left(\dfrac{4}{3} \right) =x \ln 5$ $\ln \left(\dfrac{4}{3} \right) -x \ln \left(\dfrac{4}{3} \right) = x \ln 5$ Combine the $x$ terms: $\ln \left(\dfrac{4}{3} \right) = x \ln 5 +x \ln \left(\dfrac{4}{3} \right)$ Factor out $x$ (the common factor), then solve for $x$: $\ln \left(\dfrac{4}{3} \right) = x \left(\ln 5 +\ln \left(\dfrac{4}{3} \right) \right)$ $x = \boxed{\dfrac{ \ln \left(\dfrac{4}{3} \right)}{\ln 5 +\ln \left(\dfrac{4}{3} \right)} \approx 0.152}$
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