Stitz-Zeager_College_Algebra_e-book

# A 7x1 ex1 ln7 logx 2 b log3 x 2 log3 d

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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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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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