Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Review - Exercises - Page 70: 25


(a) $$e^{2\ln 3} = 9$$ (b) $$\log_{10}25+\log_{10}4 = 2$$ (c) $$\tan\left(\arcsin \frac{1}{2}\right) = \frac{1}{\sqrt{3}}$$ (d) $$\sin\left(\cos^{-1}\frac{4}{5}\right) = \frac{3}{5}$$

Work Step by Step

(a) $$e^{2\ln 3}=e^{ln 3^2} = e^{\ln 9} = 9$$ where we used the fact that natural exponential function and natural logarithm are inverse to eachother and also the fact that $a\ln x = \ln x^a$. (b) $$\log_{10}25+\log_{10}4 = \log_{10}(25\cdot 4) = \log_{10}100 = \log_{10}10^2 =2$$ here we used the fact that the sum of two logarithms with the same base is the logarithm with that base but of the product of their arguments. Also we noted that $100=10^2$ and that the logarithm with the base $10$ is inverse of the function $10^x$. (c) We will use the trigonometric identity $$\tan x = \frac{\sin x}{\sqrt{1-\sin^2 x}} \quad x\in(-\pi/2,\pi/2).$$ $$\tan\left(\arcsin \frac{1}{2}\right) = \frac{\sin \arcsin \frac{1}{2}}{\sqrt{1-\sin^2\arcsin \frac{1}{2}}} = \frac{\frac{1}{2}}{\sqrt{1-\left(\frac{1}{2}\right)^2}} = \frac{\frac{1}{2}}{\sqrt{1-\frac{1}{3}}} = \frac{\frac{1}{2}}{\sqrt{\frac{3}{4}}} = \frac{1}{\sqrt{3}}$$ (d) We will use the trigonometric identity $$\sin x=\sqrt{1-\cos^2 x},\quad x\in(0,\pi).$$ $$\sin\left(\cos^{-1}\frac{4}{5}\right) = \sqrt{1-\cos^2\cos^{-1}\frac{4}{5}} = \sqrt{1-\left(\frac{4}{5}\right)^2} = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}}=\frac{3}{5}.$$
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