College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3: 87

Answer

$$\frac{1}{2}(A-3C)$$

Work Step by Step

$\log_b2=A$ and $\log_b3=C$ $$X=\log_b\sqrt{\frac{2}{27}}$$ $$X=\log_b\Bigg(\frac{2}{27}\Bigg)^{1/2}$$ Apply the Power Rule here, we can move the exponent $1/2$ away for Quotient Rule usage later. $$X=\frac{1}{2}\log_b\frac{2}{27}$$ Now we can apply Quotient Rule for $\log_b\frac{2}{27}$ $$X=\frac{1}{2}(\log_b2-\log_b27)$$ $$X=\frac{1}{2}(\log_b2-\log_b3^3)$$ Again, use the Power Rule for $\log_b3^3$ $$X=\frac{1}{2}(\log_b2-3\log_b3)$$ Now we substitute A and C into X $$X=\frac{1}{2}(A-3C)$$
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