Stitz-Zeager_College_Algebra_e-book

1 the symbol pka is dened similarly to ph in that pka

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Unformatted text preview: In a way, this says everything, but at the same time, it doesn’t. For example, if we try to find log2 (−1), we are trying to find the exponent we put on 2 to give us −1. In other words, we are looking for x that satisfies 2x = −1. There is no such real number, since all powers of 2 are positive. While what we have said is exactly the same thing as saying ‘the domain of g (x) = log2 (x) is (0, ∞) because the range of f (x) = 2x is (0, ∞)’, we feel it is in a student’s best interest to understand the statements in Theorem 6.2 at this level instead of just merely memorizing the facts. Example 6.1.3. Simplify the following. 1. log3 (81) 2. log2 4. ln √ 3 e2 5. log(0.001) 1 8 6. 2log2 (8) 7. 117− log117 (6) 3. log√5 (25) Solution. 1. The number log3 (81) is the exponent we put on 3 to get 81. As such, we want to write 81 as a power of 3. We find 81 = 34 , so that log3 (81) = 4. 2. To find log2 1 , we need rewrite 8 1 8 as a power of 2. We find 1 8 = 1 23 = 2−3 , so log2 1 8 = −3. √ 3. To determine log√5 (25), we need to express 25 as a power of 5. We know 25 = 52 , and √2 √ 22 √4 5 , so we have 25 = 5 5 . We get log√5 (25) = 4. 5= = √ √ 32 32 4. First, recall that the notation ln e means loge e...
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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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