Stitz-Zeager_College_Algebra_e-book

To check this against our previous calculations we nd

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Unformatted text preview: write 129 = 2log2 (129) . We’d then have 2x = 2log2 (129) , which means our solution is x = log2 (129). This makes sense because, after all, the definition of log2 (129) is ‘the exponent we put on 2 to get 129.’ Indeed we could have obtained this solution directly by rewriting the equation 2x = 129 in its logarithmic form log2 (129) = x. Either way, in order to get a reasonable decimal approximation to this number, we’d use the change of base formula, Theorem 6.7, to give us something more calculator friendly,1 say log2 (129) = ln(129) . Another way to arrive at this answer is as follows ln(2) 2x = 129 ln (2x ) = ln(129) x ln(2) = ln(129) ln(129) x= ln(2) Take the natural log of both sides. Power Rule ‘Taking the natural log’ of both sides is akin to squaring both sides: since f (x) = ln(x) is a function, as long as two quantities are equal, their natural logs are equal.2 Also note that we treat ln(2) as any other non-zero real number and divide it through3 to isolate the variable...
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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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