Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.3 - Logarithmic Functions - 4.3 Exercises - Page 353: 94

Answer

(a) $(e,\infty)$ (b) ) $f^{-1}(x)=e^{e^{e^x}}$

Work Step by Step

Given the function $f(x)=ln(ln(lnx))$ (a) The domain requirement is that $ln(lnx)\gt0$ which leads to $lnx\gt 1$ one more step gives $x\gt e$, so the domain of $f$ is $(e,\infty)$ (b) ) To find the inverse, first let $y=ln(ln(ln x))$, so $ln(ln x)=e^y$, which leads to $ln x=e^{e^y}$, and one more step gives $x=e^{e^{e^y}}$, switch $x,y$, we have $f^{-1}(x)=e^{e^{e^x}}$
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