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Published byJonah Cain Modified over 8 years ago

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Warm-Up

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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 Exponential Equations

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Solve by equating exponents 4 3x = 8 x+1 (2 2 ) 3x = (2 3 ) x+1 rewrite w/ same base 2 6x = 2 3x+3 6x = 3x+3 x = 1 Check → 4 3*1 = 8 1+1 64 = 64

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Your turn! 2 4x = 32 x-1 2 4x = (2 5 ) x-1 4x = 5x-5 5 = x Be sure to check your answer!!!

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When you can’t rewrite using the same base, you can solve by taking a log of both sides 2 x = 7 log 2 x = log 7 x log 2 = log 7 x = ≈ 2.807

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4 x = 15 log 4 x = log 15 x log 4 = log15 x = log15/log4 ≈ 1.953

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10 2x-3 +4 = 21 -4 -4 10 2x-3 = 17 log 10 2x-3 = log 17 (2x-3) log 10 = log 17 2x – 3 = log17/log 10 2x = (log 17/log10) +3 ≈ 2.115

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5 x+2 + 3 = 25 5 x+2 = 22 log 5 x+2 = log 22 (x+2) log 5 = log 22 x+2 = (log 22/log 5) x = (log22/log5) – 2 ≈ -.079

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

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log 3 (5x-1) = log 3 (x+7) 5x – 1 = x + 7 5x = x + 8 4x = 8 x = 2 and check log 3 (5*2-1) = log 3 (2+7) log 3 9 = log 3 9

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When you have a log on 1 side only, rewrite in exponential form b>0 & b≠1 if log b x = y, then b y = x

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

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log 2 3 + log 2 (x-7) = 3 log 2 3(x-7) = 3 log 2 (3x - 21) = 3 2 3 = 3x – 21 8 = 3x – 21 29 = 3x 29/3 = x

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log5x + log(x+1)=2 log (5x)(x+1) = 2 (product property) log (5x 2 – 5x) = 2 5x 2 - 5x = 10 2 (rewrite) 5x 2 - 5x = 100 x 2 – 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

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Homework: Page 505 –# 26, 28, 33, 35, 45, 50, 53, 56

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Answers to HW p. 505 26) x = -1/528) no sol 33) x = 3/235) x = 0.350 45) x = 218750) x = 15 53) No sol. (x = 2 but it doesn’t work when you plug it in and check) 56) No sol. (x = 2 but it doesn’t work when you plug it in and check)

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Natural Exponential Function y = e x Natural Base ln e = 1

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Natural Logarithmic Function y = e x x = log e y x = ln y

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To solve natural log functions, you solve them the same as logarithmic functions. Remember: ln has base e ln e = 1

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1.) e x+7 = 98 2.) 4e 3x-5 = 72 3.) ln x 3 - 5 = 1

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5 – ln x = 7 - ln x = 2 ln x = -2 e 2 = x 7.389 = x

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3e -x – 4 = 9 3e -x = 13 e -x = 13/3 ln e -x = ln (13/3) -x ln e = 1.466 x = -1.466

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Homework Page 505 –#32, 34, 40, 43, 46, 48

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