Stitz-Zeager_College_Algebra_e-book

Since t 9 the average rate of change of a function

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Unformatted text preview: 26 < 0, 26. √ we get no real solution to ex = 5 − 26, but for ex = 5 + 26, we take natural logs to obtain √ x −e−x x = ln 5 + 26 . If we graph f (x) = e 2 and g (x) = 5, we see that the graphs intersect √ at x = ln 5 + 26 ≈ 2.312 6.3 Exponential Equations and Inequalities 361 x −x y = f (x) = e −e and 2 y = g (x) = 5 y = f (x) = 25x and y = g (x) = 5x + 6 The authors would be remiss not to mention that Example 6.3.1 still holds great educational value. Much can be learned about logarithms and exponentials by verifying the solutions obtained in Example 6.3.1 analytically. For example, to verify our solution to 2000 = 1000 · 3−0.1t , we substitute t = − 10 ln(2) and obtain ln(3) ? −0.1 − ? 10 ln(2) ln(3) ln(2) 2000 = 1000 · 3 2000 = 1000 · 3 ln(3) ? Change of Base ? Inverse Property 2000 = 1000 · 3log3 (2) 2000 = 1000 · 2 2000 = 2000 The other solutions can be verified by using a combination of log and inverse properties. Some fall out quite quickly, while others are mor...
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