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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. - PowerPoint PPT Presentation
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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?
• 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
Solve by equating exponents
• 43x = 8x+1
• (22)3x = (23)x+1 rewrite w/ same base
• 26x = 23x+3
• 6x = 3x+3
• x = 1Check → 43*1 = 81+1
64 = 64
Your turn!
• 24x = 32x-1
• 24x = (25)x-1
• 4x = 5x-5
• 5 = x
Be sure to check your answer!!!
Solve the Equation1. 9 = 27 2x x – 1
SOLUTION
Rewrite 9 and 27 as powers with base 3.
Write original equation.
Power of a power propertyProperty of equality for exponential equations
4x – 3x = –3
= –3Solve for x.
Property of equality for exponential equations
The solution is – 3.
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
Use log2 because the x is on the 2 and log22=1
4x = 15• log44x = log415
• x = log415 = log15/log4
• ≈ 1.95
Use change of base to solve
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
5x+2 + 3 = 25• 5x+2 = 22• log55x+2 = log522• x+2 = log522• x = (log522) – 2• = (log22/log5) – 2• ≈ -.079
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
• 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°?
• 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?
• 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.
• 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.
4.6 Assignment
Page 271, 5-10, 14-21, 54-58
Solve Exponential and Logarithmic
Equations 4.6Day 2
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
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
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
5x – 1 = 64
5x = 65
x = 13
SOLUTION
Write original equation.
Exponentiate each side using base 4.
Add 1 to each side.
Divide each side by 5.
Solve (5x – 1)= 3log4
4log4
(5x – 1) = 43
(5x – 1)= (5x – 1)= 3log4
b = xlogbx
The solution is 13.ANSWER
This is the way the book suggests you do the problem.
Solve (5x – 1)= 3log4
Solve using the definition
Use the definition
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
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
Solve the equation. Check for extraneous solutions.
ln (7x – 4) = ln (2x + 11)
SOLUTION
Write original equation.ln (7x – 4) = ln (2x + 11)
7x – 4 = 2x + 11
7x – 2x = 11 – 4
5x = 15
x = 3
The solution is 3.ANSWER
Property of equality for logarithmic equations
Divide each side by 5.
log 5x + log (x – 1) = 2
Solve the equation. Check for extraneous solutions.
SOLUTION
log 5x + log (x – 5) = 2
log [5x(x – 1)] = 2
5x(x – 1) = 100
Write original equation.
Product property of logarithms
Use the definition
Distributive property
Subtract 100
Divide out a 5
FactorZero product property
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
Assignment 4.6 day 2
• p. 271, 26-42 all