**Unformatted text preview: **, so we are looking for the exponent
√
√
√
3
3
32
to put on e to obtain e2 . Rewriting e2 = e2/3 , we ﬁnd ln
e = ln e2/3 = 2 .
3
5. Rewriting log(0.001) as log10 (0.001), we see that we need to write 0.001 as a power of 10. We
1
1
have 0.001 = 1000 = 103 = 10−3 . Hence, log(0.001) = log 10−3 = −3.
6. We can use Theorem 6.2 directly to simplify 2log2 (8) = 8. We can also understand this problem
by ﬁrst ﬁnding log2 (8). By deﬁnition, log2 (8) is the exponent we put on 2 to get 8. Since
8 = 23 , we have log2 (8) = 3. We now substitute to ﬁnd 2log2 (8) = 23 = 8.
7. We note that we cannot apply Theorem 6.2 directly to 117− log117 (6) . (Why not?) We use
1
a property of exponents to rewrite 117− log117 (6) as
. At this point, we can apply
117log117 (6) 6.1 Introduction to Exponential and Logarithmic Functions
Theorem 6.2 to get 117log117 (6) = 6 and thus 117− log117 (6) = 337
1 117log117 (6)
117log117 (6) = = 1
6. It is worth a moment of your time to think your way through why
6. By deﬁnition,
log117 (6) is the exponent we put on 117...

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