Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: og(x) y= 1 − log(x) log(y ) x= Interchange x and y . 1 − log(y ) x (1 − log(y )) = log(y ) x − x log(y ) = log(y ) x = x log(y ) + log(y ) x = (x + 1) log(y ) x = log(y ) x+1 x y = 10 x+1 Rewrite as an exponential equation. 3 Refer to page 158 for a discussion of what this means. 374 Exponential and Logarithmic Functions x We have f −1 (x) = 10 x+1 . Graphing f and f −1 on the same viewing window yields y = f (x) = x log(x) and y = g (x) = 10 x+1 1 − log(x) 6.4 Logarithmic Equations and Inequalities 6.4.1 375 Exercises 1. Solve the following equations analytically. x = 150 10−12 (j) log3 (x) = log 1 (x) + 8 (a) log 1 x = −3 (i) 10 log 2 (b) ln(x2 ) = (ln(x))2 3 (c) log3 (x − 4) + log3 (x + 4) = 2 (d) log5 (2x + 1) + log5 (x + 2) = 1 (e) log2 (x3 ) = log2 (x) (f) log169 (3x + 7) − log169 (5x − 9) = x (g) log = 4.7 10−3 (h) − log(x) = 5.4 1 2 1 3x − 2 = (k) log125 2x + 3 3 (l) ln(x + 1) − ln(x) = 3 (m) ln(ln(x)) = 3 (n) 2 log7 (x) = log7 (2) + log7 (x + 12) (o)...
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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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