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 351: 7

Answer

$$\log_8{8}=1$$ $$\underline{8^1=8}$$ -- $$log_8{64}=2$$ $$\underline{8^2=64}$$ -- $$\underline{8^{\frac{2}{3}}=4}$$ $$log_8{4}=\frac{2}{3}$$ -- $$\underline{8^3=512}$$ $$log_8{512}=3$$ -- $$log_8{\frac{1}{8}}=-1$$ $$\underline{8^{-1}=\frac{1}{8}}$$ -- $$\underline{8^{-2}=\frac{1}{64}}$$ $$log_8{\frac{1}{64}}=-2$$

Work Step by Step

Before we begin, let's shortly overview what is logarithmic expression. $log_{a}{x}=b$ this is a logarithmic form which by exponential form means $a^b=x$. According to this we can easily fill the table. $\log_8{8}=1$ This simply means: $8^1=8$ $log_8{64}=2$ It means: $8^2=64$ This time we have vice versa: $8^{\frac{2}{3}}=4$ And it means: $log_8{4}=\frac{2}{3}$ $8^3=512$ It stands for: $log_8{512}=3$ $log_8{\frac{1}{8}}=-1$ In this case: $8^{-1}=\frac{1}{8}$ And the last one: $8^{-2}=\frac{1}{64}$ It is similar as: $log_8{\frac{1}{64}}=-2$
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