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L5.7 Exponential Equations and Changing Bases

L5.7 Exponential Equations and Changing Bases. If f(x) = log 3 x, show that. L5.7 Warm up. The above shows that horizontally stretching the graph of f by 3 is equivalent to. shifting down one unit. Is this true for all log functions?.

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L5.7 Exponential Equations and Changing Bases

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  1. L5.7 Exponential Equations and Changing Bases

  2. If f(x) = log3 x, show that L5.7 Warm up The above shows that horizontally stretching the graph of f by 3 is equivalent to . shifting down one unit Is this true for all log functions? For f(x) = logb x, horizontally stretching the graph by b is equivalent to shifting down one unit.

  3. Exponential Equations An exponential equation is an equation that contains a variable in the exponent. To solve, you must get the variable out of the exponent. There are two ways to solve: • If you can rewrite the two sides in terms of the same base, you can equate exponents: bx = by↔ x = y • Otherwise you take the log* of both sides and pull the exponent in front, using the Power of Logs Law from L5.6 (i.e., logbxk = k·logbx). • * Use the common log or natural log so you can solve using a calculator.

  4. Exponential Inequalities Like logarithmic inequalities, whether the base is a fraction or not determines whether the inequality holds in the current direction or must be reversed. • If b > 1, then bx > by↔ x > y • If 0 < b < 1, then bx > by ↔ x < y. DO PROBLEMS ON HANDOUT

  5. Example Problems (1 of 3) The population of a town is currently 50,000 and is growing at an annual rate of 2.5%. The growth in terms of number of years from now is given by P(t) = 50,000 (1.025)t. At this rate, how long will it take for the population to reach 120,000? 120,000 = 50,000(1.025)t 2.4 = 1.025t ln(2.4) = ln(1.025)t ln(2.4) = t·ln(1.025) t = ln(2.4) / ln(1.025) t ≈ 35.5 years Do NOT estimate as you go. Wait until the end.

  6. Example Problems (2 of 3) Suppose you invest P dollars at an annual rate of 6% compounded daily. How long will it take to (a) increase the investment by 50%? (b) triple your money? For this example, assume compounding daily ~ continuous compounding, so use P(t) = Pert a) 1.5P = Pe.06t 1.5 = e.06t ln 1.5 = ln e.06t ln 1.5 = .06t t = ln 1.5 / .06 ≈ 6.76 years or 6 yrs, 9 mos b) 3P = Pe.06t 3 = e.06t ln 3 = ln e.06t ln 3 = .06t t = ln 3 / .06 ≈ 18.31 years, or 18 yrs, 4 mos

  7. Example Problems (3 of 3) Find the x-intercept of f(x) = 23x–2 – 5. 23x–2 – 5 = 0 23x–2 = 5 ln(23x–2) = ln 5 (3x – 2) ln 2 = ln 5 3x – 2 = ln 5 / ln 2 3x = (ln 5 / ln 2) + 2 x = ⅓[(ln 5 / ln 2) + 2] ≈ 1.4406

  8. Change of Base Formula Thus, Most calculators have 2 logs built in: base 10 and base e. The Change of Base Formula allows you to find the log of any base. a can be any base, but generally a = 10 or a = e, so a calculator can be used. Find log5 290. Using ln generates the same ans. Derivation of the Change of Base Formula: Let x = logbc bx = c loga bx = loga c x loga b = loga c

  9. Homework Day 1: pp 205-207 #3, 7, 8, 11, 13, 14, 17, 18, 27, 29, 32 Day 2: p207 #35, 36, 37, 39, 41, 42, 44; p179 #43-48 all

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