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4.6 Solve Exponential and Logarithmic Equations

4.6 Solve Exponential and Logarithmic Equations. p. 267 How do you use logs to solve an exponential equation? When is it easiest to use the definition of logs? Do you ever get a negative answer for logs?. Exponential Equations.

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4.6 Solve Exponential and Logarithmic Equations

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  1. 4.6 Solve Exponential and Logarithmic Equations p. 267 How do you use logs to solve an exponential equation? When is it easiest to use the definition of logs? Do you ever get a negative answer for logs?

  2. Exponential Equations • One way to solve exponential equations is to use the property that if 2 powers w/ the same base are equal, then their exponents are equal. • For b>0 & b≠1 if bx = by, then x=y

  3. Solve by equating exponents • 43x = 8x+1 • (22)3x = (23)x+1 rewrite w/ same base • 26x = 23x+3 • 6x = 3x+3 • x = 1 Check → 43*1 = 81+1 64 = 64

  4. Your turn! • 24x = 32x-1 • 24x = (25)x-1 • 4x = 5x-5 • 5 = x Be sure to check your answer!!!

  5. x –1 2x 1. 9 = 27 Solve the Equation SOLUTION Write original equation. Rewrite 9 and 27 as powers with base 3. Power of a power property Property of equality for exponential equations 4x – 3x =–3 Property of equality for exponential equations =–3 Solve for x. The solution is – 3.

  6. When you can’t rewrite using the same base, you can solve by taking a log of both sides • 2x = 7 • log22x = log27 • x = log27 • x = ≈ 2.807 Use log2 because the x is on the 2 and log22=1

  7. 4x = 15 • log44x = log415 • x = log415 = log15/log4 • ≈ 1.95 Use change of base to solve

  8. 102x-3+4 = 21 • -4 -4 • 102x-3 = 17 • log10102x-3 = log1017 • 2x-3 = log 17 • 2x = 3 + log17 • x = ½(3 + log17) • ≈ 2.115

  9. 5x+2 + 3 = 25 • 5x+2 = 22 • log55x+2 = log522 • x+2 = log522 • x = (log522) – 2 • = (log22/log5) – 2 • ≈ -.079

  10. Newton’s Law of Cooling • The temperature T of a cooling substance @ time t (in minutes) is: • T = (T0 – TR) e-rt + TR • T0= initial temperature • TR= room temperature • r = constant cooling rate of the substance

  11. You’re cooking stew. When you take it off the stove the temp. is 212°F. The room temp. is 70°F and the cooling rate of the stew is r =.046. How long will it take to cool the stew to a serving temp. of 100°?

  12. T0 = 212, TR = 70, T = 100 r = .046 • So solve: • 100 = (212 – 70)e-.046t +70 • 30 = 142e-.046t (subtract 70) • .221 ≈ e-.046t (divide by 142) • How do you get the variable out of the exponent?

  13. Cooling cont. • ln .221 ≈ ln e-.046t(take the ln of both sides) • ln .221 ≈ -.046t • -1.556 ≈ -.046t • 33.8 ≈ t • about 34 minutes to cool!

  14. How do you use logs to solve an exponential equation? Expand the logs to bring the exponent x down and solve for x. • When is it easiest to use the definition of logs? When you have log information on the left equal to a number on the right. • Do you ever get a negative answer for logs? Never! Logs are always positive.

  15. 4.6 Assignment Page 271, 5-10, 14-21, 54-58

  16. Solve Exponential and Logarithmic Equations 4.6Day 2

  17. Solving Log Equations • To solve use the property for logs w/ the same base: • Positive numbers b,x,y & b≠1 • If logbx = logby, then x = y

  18. log3(5x-1) = log3(x+7) • 5x – 1 = x + 7 • 5x = x + 8 • 4x = 8 • x = 2 and check • log3(5*2-1) = log3(2+7) • log39 = log39

  19. When you can’t rewrite both sides as logs w/ the same base exponentiate each side • b>0 & b≠1 • if x = y, then bx = by

  20. Solve (5x – 1)= 3 (5x – 1)= 3 (5x – 1)= 3 4log4(5x – 1) = 4 log log log 4 4 b ANSWER x The solution is 13. b = x SOLUTION Write original equation. Exponentiate each side using base 4. 5x – 1 = 64 5x = 65 Add 1 to each side. x = 13 Divide each side by 5. This is the way the book suggests you do the problem.

  21. Solve (5x – 1)= 3 log 4 Solve using the definition Use the definition

  22. log5(3x + 1) = 2 • 52 = (3x+1) (use definition) • 3x+1 = 25 • x = 8 and check • Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions

  23. log5x + log(x-1)=2 • log (5x)(x-1) = 2 (product property) • log (5x2 – 5x) = 2 (use definition) • 5x2−5x = 102 • 5x2 - 5x = 100 • x2 – x - 20 = 0 (subtract 100 and divide by 5) • (x-5)(x+4) = 0 x=5, x=-4 • graph and you’ll see 5=x is the only solution 2

  24. ANSWER The solution is 3. Solve the equation. Check for extraneous solutions. ln (7x – 4) = ln (2x + 11) SOLUTION ln (7x – 4) = ln (2x + 11) Write original equation. 7x – 4 = 2x + 11 Property of equality for logarithmic equations 7x – 2x = 11 – 4 5x = 15 Divide each side by 5. x = 3

  25. Solve the equation. Check for extraneous solutions. log 5x + log (x – 1) = 2 SOLUTION log 5x +log(x – 5) = 2 Write original equation. log [5x(x – 1)] = 2 Product property of logarithms Use the definition Distributive property 5x(x – 1) = 100 Subtract 100 Divide out a 5 Factor Zero product property

  26. One More!log2x + log2(x-7) = 3 • log2x(x-7) = 3 • log2 (x2- 7x) = 3 • x2−7x = 23 • x2 – 7x = 8 • x2 – 7x – 8 = 0 • (x-8)(x+1)=0 • x=8 x= -1 2

  27. Assignment 4.6 day 2 • p. 271, 26-42 all

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