Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Exercises - Page 389: 49


$2\log_5 x$ + $\frac{3}{2}\log_5 (1-5x)$ - $\frac{1}{2}\log_5 (x^3-x)$

Work Step by Step

$Expand$ $the$ $logarithmic$ $expression:$ $\log_5 (\frac{x^2(1-5x)^{\frac{3}{2}}}{\sqrt {x^3-x}})$ Use the Second Law of logarithms $\log_5 (\frac{x^2(1-5x)^{\frac{3}{2}}}{\sqrt {x^3-x}})$ = $\log_5 (x^2 (1-5x)^{\frac{3}{2}})$ - $\log_5 \sqrt {x^3 - x}$ Use the First Law of Logarithms for $\log_5 (x^2 (1-5x)^{\frac{3}{2}})$ $\log_5 (x^2 \times(1-5x)^{\frac{3}{2}})$ = $\log_5 x^2$ + $\log_5 (1-5x)^{\frac{3}{2}}$ $\log_5 x^2$ + $\log_5 (1-5x)^{\frac{3}{2}}$ - $\log_5 \sqrt{x^3-x}$ Rewrite the root to exponent form $\log_5 x^2$ + $\log_5 (1-5x)^{\frac{3}{2}}$ - $\log_5 (x^3-x)^{\frac{1}{2}}$ Use the Third Law of Logarithms for the whole expression $\log_5 x^2 = 2\log_5 x$ $\log_5 (1-5x)^{\frac{3}{2}}$ = $\frac{3}{2}\log_5 (1-5x)$ $\log_5 (x^3-x)^{\frac{1}{2}}$ = $\frac{1}{2}\log_5 (x^3-x)$ $2\log_5 x$ + $\frac{3}{2}\log_5 (1-5x)$ - $\frac{1}{2}\log_5 (x^3-x)$
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