Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 9 - Section 9.3 - Logarithmic Functions and Models - Exercises - Page 656: 2

Answer

1) The exponential form: $4^{3}=64$ The logarithmic form: $\log_{4}64=3$. 2) The exponential form: $10^{-1}=0.1$ The logarithmic form: $\log_{10}0.1=-1$. 3) The exponential form: $2^{8}=256$ The logarithmic form: $\log_{2}256=8$. 4) The exponential form: $5^{0}=1$ The logarithmic form: $\log_{5}1=0$. 5) The exponential form: $0.5^{2}=0.25$ The logarithmic form: $\log_{0.5}0.25=2$. 6) The exponential form: $6^{-2}=\frac{1}{36}$ The logarithmic form: $\log_{6}\frac{1}{36}=-2$

Work Step by Step

The definition of the logarithm function can be translated into mathematical formulas such as the following. The given expressions are equivalent. The exponential form: $b^{x}=a$ The logarithmic form: $\log_{b}a=x$. Here, we have to translate the given expressions to the logarithmic form: 1) The exponential form: $4^{3}=64$ The logarithmic form: $\log_{4}64=3$. 2) The exponential form: $10^{-1}=0.1$ The logarithmic form: $\log_{10}0.1=-1$. 3) The exponential form: $2^{8}=256$ The logarithmic form: $\log_{2}256=8$. 4) The exponential form: $5^{0}=1$ The logarithmic form: $\log_{5}1=0$. 5) The exponential form: $0.5^{2}=0.25$ The logarithmic form: $\log_{0.5}0.25=2$. 6) The exponential form: $6^{-2}=\frac{1}{36}$ The logarithmic form: $\log_{6}\frac{1}{36}=-2$
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