5.6 - Solving Logarithmic and Exponential Equations Exponentials: ln(ex) = x… so if ex = 2… ln(ex) = ln(2) x = ln(2) ex: ex + 5 = 60 ex = 55 ln(ex) = ln(55) x = ln(55) ex: 4e2x = 5 e2x = 5/4 ln(e2x) = ln(5/4) 2x = ln(5/4) x = ln(5/4) / 2

ex: e2x - 3ex + 2 = 0 (ex - 2) (ex - 1) = 0 ex: Other Bases 2x = 10 ln(2x) = ln(10) x ln(2) = ln(10) x = ln(10) / ln(2) = 3.322 (ex - 2) = 0 ex = 2 ln(ex) = ln(2) x = ln(2) = 0.693 (ex - 1) = 0 ex = 1 ln(ex) = ln(1) x = 0

Finding Solutions Graphically 2) Set equation = 0 Graph Find zero(s) ex: 2x = 10 3) 1)

IC 5.6A - pg 386 #’s 1, 2, 4, 5, 8, 10, 11, 14, 15, 18, 19, 22, 25, 26, 30, 31, 34.

e(ln 3x) = e2 Solving Logarithmic Equations Exponentials and logs are inverse functions So for logs we raise both sides of the equation up as exponents…the base of the exponential we are creating is the base of the log statement (e for ln’s / whatever ‘a’ is for logs) We call this ‘exponentiation’ ex: ln x = 3 eln x = e3 x = e3 5 + 2ln x = 4 2ln x = -1 ln x = -1/2 eln x = e(-1/2) x = e(-1/2) 2ln 3x = 4 ln 3x = 2 e(ln 3x) = e2 3x = e2 x = e2/3

2(log23x) = 22 = 64 Other Bases ex: log4x = 3 4(log4x) = 43 x = 43

Shortcut ex: log10(x + 2) = log10 (3x - 4) Drop log statements (x + 2) = (3x - 4) 2x = 6 x = 3 Multiple log (or ln) statements ln(x - 3) + ln(2x + 1) = 2(ln x) Combine with laws of logs ln[(x – 3)(2x + 1)] = ln x2 Drop log statements (x – 3)(2x + 1) = x2 2x2 - 5x – 3 = x2 x2 - 5x – 3 = 0 Factor / Graph

Creating log (or ln) statements from constants loga ax = x so… log10 10 = 1 ln e = 1 ln e2 = 2 ln e3 = 3 log10 102 = 2 log10 103 = 3 etc… ex: 2 + ln (x + 4) = ln 16 ln e2 + ln (x + 4) = ln 16 ln e2 (x + 4) = ln 16 Combine log statements e2 (x + 4) = 16 Drop log statements x = (16/e2) - 4

IC 5.6B – pg 386 #’s 37, 38, 41, 42, 45-48 all, 52.