University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 50: 65


(a) $\frac{\log_2x}{\log_3x}=\log_23$ (b) $\frac{\log_2x}{\log_8x}=3$ (c) $\frac{\log_xa }{\log_{x^2}a}=2$

Work Step by Step

*Recall the Change of Base Formula: $$\log_ax=\frac{\ln x}{\ln a}$$ (a) $$\frac{\log_2x}{\log_3x}$$ Apply the Change of Base Formula here: $\log_2x=\frac{\ln x}{\ln 2}$ and $\log_3x=\frac{\ln x}{\ln 3}$ Therefore, $$\frac{\log_2x}{\log_3x}=\frac{\frac{\ln x}{\ln 2}}{\frac{\ln x}{\ln3}}=\frac{\ln x}{\ln 2}\times\frac{\ln3}{\ln x}=\frac{\ln 3}{\ln2}$$ Here we can apply the Change of Base Formula one more time, meaning $\frac{\ln 3}{\ln2}=\log_23$ In conclusion: $$\frac{\log_2x}{\log_3x}=\log_23$$ (b) $$\frac{\log_2x}{\log_8x}$$ Apply the Change of Base Formula here: $\log_2x=\frac{\ln x}{\ln 2}$ and $\log_8x=\frac{\ln x}{\ln 8}$ Therefore, $$\frac{\log_2x}{\log_8x}=\frac{\frac{\ln x}{\ln 2}}{\frac{\ln x}{\ln8}}=\frac{\ln x}{\ln 2}\times\frac{\ln8}{\ln x}=\frac{\ln 8}{\ln2}=\frac{\ln2^3}{\ln2}$$ Apply Power Rule: $$\frac{\log_2x}{\log_8x}=\frac{3\ln2}{\ln2}=3$$ (c) $$\frac{\log_xa }{\log_{x^2}a}$$ Apply the Change of Base Formula here: $\log_xa=\frac{\ln a}{\ln x}$ and $\log_{x^2}a=\frac{\ln a}{\ln x^2}$ Therefore, $$\frac{\log_xa}{\log_{x^2}a}=\frac{\frac{\ln a}{\ln x}}{\frac{\ln a}{\ln x^2}}=\frac{\ln a}{\ln x}\times\frac{\ln x^2}{\ln a}=\frac{\ln x^2}{\ln x}$$ Apply Power Rule: $$\frac{\log_xa}{\log_{x^2}a}=\frac{2\ln x}{\ln x}=2$$
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