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: 64


(a) $25^{\log_53x^2}=9x^4$ (b) $\log_e(e^x)=x$ (c) $\log_4(2^{e^x\sin x})=\frac{1}{2}e^x\sin x$

Work Step by Step

(a) $$25^{\log_53x^2}$$ Here $25$ and $5$ do not match each other to use the inverse properties. But $25=5^2$ That means $$25^{\log_53x^2}=5^{2\log_53x^2}$$ - Apply Power Rule, we have: $2\log_53x^2=\log_5(3x^2)^2=\log_59x^4$ Thus, $$25^{\log_53x^2}=5^{\log_59x^4}=9x^4$$ (b) $$\log_e(e^x)$$ We can apply right away the inverse properties here: $$\log_e(e^x)=x$$ (c) $$\log_4(2^{e^x\sin x})$$ Notice that $2=\sqrt4=4^{\frac{1}{2}}$ So we can rewrite $2^{e^x\sin x}=4^{\frac{1}{2}e^x\sin x}$ Therefore, $$\log_4(2^{e^x\sin x})=\log_44^{\frac{1}{2}e^x\sin x}$$ Now apply the inverse property with base $a=4$: $$\log_4(2^{e^x\sin x})=\frac{1}{2}e^x\sin x$$
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