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 49: 46


(a) $e^{\ln(x^2+y^2)}= x^2+y^2$ (b) $e^{-\ln 0.3}=\frac{10}{3}$ (c) $e^{\ln \pi x-\ln 2}=\frac{\pi x}{2}$

Work Step by Step

This whole exercise is based on this property: $$e^{\ln x}=x\hspace{1cm}x\gt0$$ (a) $$e^{\ln(x^2+y^2)}= x^2+y^2$$ (b) $$e^{-\ln 0.3}$$ - Apply Reciprocal Rule for $-\ln 0.3$: $-\ln 0.3=\ln\frac{1}{0.3}=\ln\frac{10}{3}$ $$e^{-\ln 0.3}=e^{\ln\frac{10}{3}}=\frac{10}{3}$$ (c) $$e^{\ln \pi x-\ln 2}$$ - Apply Quotient Rule here: $\ln \pi x-\ln 2 =\ln\frac{\pi x}{2}$ $$e^{\ln \pi x-\ln 2}=e^{\ln\frac{\pi x}{2}}=\frac{\pi x}{2}$$
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