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: 48


$\text{Exact: } \dfrac{5 \ln \left(\dfrac{2}{3} \right)}{ \ln 4}$ $\text{Approximately: } -1.462$

Work Step by Step

Divide both sides of the equation by 0.3: $$4^{0.2x}= \dfrac{0.2}{0.3} $$ $$4^{0.2x} = \dfrac{2}{3}$$ $\because a^y = b \text{ is equivalent to } y = \log_a b$ $\therefore 4^{0.2x} = \dfrac{2}{3} \text{ is equivalent to }0.2x = \log_4 \left(\dfrac{2}{3} \right)$ SOolve the equation above using the Change of Base Formula, which is $\hspace{20pt} \log_ a M = \dfrac{\log_b M}{\log_b a}$, to obtain: $0.2x=\log_4 \left(\dfrac{2}{3} \right)\\\\ 5(0.2x) = 5\left(\dfrac{\ln \left(\dfrac{2}{3} \right)}{\ln 4}\right)\\\\ x = \dfrac{5\ln \left(\dfrac{2}{3} \right)}{\ln 4} $ Therefore, $x =\boxed{ \dfrac{5 \ln \left(\dfrac{2}{3} \right)}{ \ln 4} \approx -1.462}$
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