8.6 solving exponential and logarithmic equations
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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. - PowerPoint PPT PresentationTRANSCRIPT
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8.6Solving Exponential and Logarithmic Equations
p. 501
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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 bx = by, then x=y
Exponential Equations
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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
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Your turn!
• 24x = 32x-1
• 24x = (25)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
• 2x = 7
• log22x = log27
• x = log27
• x = ≈ 2.8072log
7log
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4x = 15• log44x = log415
• x = log415 = log15/log4
• ≈ 1.95
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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
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5x+2 + 3 = 25• 5x+2 = 22• log55x+2 = log522• x+2 = log522• x = (log522) – 2• = (log22/log5) – 2• ≈ -.079
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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
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• 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°?
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• 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?
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• 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!
Cooling cont.
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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 logbx = logby, then x = y
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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
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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
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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
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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
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One More!
log2x + log2(x-7) = 3• log2x(x-7) = 3• log2 (x2- 7x) = 3• 2log
2x -7x = 32
• x2 – 7x = 8• x2 – 7x – 8 = 0• (x-8)(x+1)=0• x=8 x= -1
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