Calculus: Early Transcendentals 8th Edition

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

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 67: 53


a) $x=5+\log_23$ b) $x=\frac{1}{2}(1+\sqrt{4e+1})$

Work Step by Step

a) $2^{x-5}=3$ $x-5=\log_23$ $x=5+\log_23$ b) $\ln x+\ln (x-1)=1$, $x\gt1$ $\ln x(x-1)=1$ $x(x-1)=e$ $x^2-x=e$ $x^2-x+\frac{1}{4}=e+\frac{1}{4}$ $(x-\frac{1}{2})^2=e+\frac{1}{4}$ $x-\frac{1}{2}=\pm\sqrt{e+\frac{1}{4}}$ $x=\frac{1}{2}\pm\sqrt{\frac{1}{4}(4e+1)}\gt0$ $\therefore x=\frac{1}{2}+\frac{1}{2}\sqrt{4e+1}$ $=\frac{1}{2}(1+\sqrt{4e+1})$
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