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8.6 Solving Exponential and Logarithmic Equations

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19 diapositives · 374 Views · Aug 10, 2014

8.6 Solving Exponential and Logarithmic Equations. p. 501. 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 b x = b y , then x=y.

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8.6 Solving Exponential and Logarithmic Equations
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  1. 8.6Solving Exponential and Logarithmic Equations p. 501

  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. 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

  6. 4x = 15 • log44x = log415 • x = log415 = log15/log4 • ≈ 1.95

  7. 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

  8. 5x+2 + 3 = 25 • 5x+2 = 22 • log55x+2 = log522 • x+2 = log522 • x = (log522) – 2 • = (log22/log5) – 2 • ≈ -.079

  9. 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

  10. 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°?

  11. 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?

  12. 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!

  13. Solving Log Equations • To solve use the property for logs w/ the same base: • + #’s b,x,y & b≠1 • If logbx = logby, then x = y

  14. 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

  15. 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

  16. log5(3x + 1) = 2 • 5log5(3x+1) = 52 • 3x+1 = 25 • x = 8 and check • Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions

  17. log5x + log(x+1)=2 • log (5x)(x+1) = 2 (product property) • log (5x2 – 5x) = 2 • 10log5x -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

  18. One More!log2x + log2(x-7) = 3 • log2x(x-7) = 3 • log2 (x2- 7x) = 3 • 2log2x -7x = 32 • x2 – 7x = 8 • x2 – 7x – 8 = 0 • (x-8)(x+1)=0 • x=8 x= -1 2

  19. Assignment