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: 14



Work Step by Step

The domain of the variable requires that $x>0$. Recall: $\log_a M^r = r \log_a M$ This means that: $2\log_5 x = \log_5 x^2\\$ $3 \log_5 4 = \log_5 4^3$ Thus, the given equation is equivalent to: $\log_5 x^2 = \log_5 4^3$ Recall also that: $\text{If } \log_a M = \log_a N \text{, then } M=N$. Therefore, $x^2 = 4^3$ $x^2 = 64$ $x = \pm \sqrt{64}$ $x = \pm 8$ $x$ cannot be negative so $x=-8$ is not a solution. Hence, $x = \boxed{8}$
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