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Solving Exponential and Logarithmic Equations
Solving when the exponent has a variable.
Rules of exponents
• Multiply……add
• Divide………subtract
• Power to a power….multiply
x0 =1
x−1 =1x
x =x12
Radicals become Fractional exponents:
power
rootx23 =x
23
Practice with exponents
• See textbook pg. 292
Radical expressions
• Express the when coverting a radical
• Expression with a fractional exponent.
• Example:
• 1. What is the power?
• 2. what is the root?
power
root
835
Fractional exponents
• Express as
•
power
root
835= 8
53
Now evaluate:
= 32
Practice:
• See page 296 in textbook.do 3-6
• 297 20, 21 46-48, 58, 68,69
• Do these:1. 44
2
2. 73
2
Do now:
• Express with fractional exponents:
• Solve:
x + 3
x + 3 =2
Solving when the base is a variable:
• Solve the previous problem:
(x + 3)12 =2
((x+ 3)12 )2 =22
x+ 3=4x=1
Procedure:
• Isolate the exponential, raise to the reciprocal power, and solve:
4x3
2 =108
Procedure:
• Isolate the exponential, raise to the reciprocal power, and solve:
4x3
2 =108
x32 =27
(x32 )
23 =27
23
x=9
Example:
• Isolate first!
3x3 −2 =22
Example:
• Isolate first!
3x3 −2 =22
3x3 =24
x3 =8
x=813 =2
• 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.
• if bx = by, then x=y• For b>0 & b≠1
Exponential Equations
base 2, base 3,base 5 numbers:
• base 2: base 3: base 5:
• 2 3 5
• 4 9 25
• 8 27 125
• 16 81
• 32
• 64
Solve by equating exponents
• 43x = 8x+1
• But 4 = 22 and 8 = 23
• (22)3x = (23)x+1 rewrite w/ same base
Solve by equating exponents
• 43x = 8x+1
• (22)3x = (23)x+1 rewrite w/ same base
• 26x = 23x+3 power to a power: multiply
• 6x = 3x+3 set exponents equal
• x = 1 Check → 43*1 = 81+1
64 = 64
Procedure:
• Copy these steps
• 1. decide like base
• 2. rewrite one or both numbers as a power of the base
• 3. simplify the exponents
• 4. set exponents equal and solve
• 5. be sure to check!
Your turn!
• 24x = 32x-1
Solution
• 24x = 32x-1
• 24x = (25)x-1
• 4x = 5x-5
• 5 = x
Be sure to check your answer!!!
Hand this one in:
• Solve and check:
34 x =27x+1
Hand this one in:
• Solve and check:
Graphs of exponentials
• Growth and decay:
• growth decay
Interest rate formula:
• A=a0(1+r)t
Exponential growth and decay formula:
y = abx
a is the initial value b is used to find the growth or decay rate.x is the time If b >1 then b - 1 is the growth rateIf b < 1 then 1 – b is the decay rate
Note: the rate is always in decimal form.
Interest rate
• Use the formula to determine the amount of money you will have after 3 years if you invest 100 at a rate of 8%.
• A=a0(1+r)t
Interest rate
• Use the formula to determine the amount of money you will have after 3 years if you invest 100 at a rate of 8%.
a =a0 (1+ r)t
a=100(1+.08)t
a=100(1.08)3
a=125.97
Exponential examples
• Y = 300(.95)x
• What is the initial value?
• What is the rate of decay?
• How much will there be in 10 years?
• Y = 20(1.3)t
• What is the initial value
• What is the growth rate?
• How much will it be in 5 years?
COMPOUND INTEREST FORMULA
amount at the end
Principal(amount at start)
annual interest rate
(as a decimal)
A =a0 1+rn
⎛
⎝⎜
⎞
⎠⎟
nttime(in
years)
number of times per year that interest in
compounded
Calculate how much you will have ifYou invest 100 at 8% compounded Quarterly after 3 years.
A =a0 1+rn
⎛
⎝⎜
⎞
⎠⎟
nt
Calculate how much you will have ifYou invest 100 at 8% compounded Quarterly after 3 years.
A =a0 1+rn
⎛
⎝⎜
⎞
⎠⎟
nt
a=100(1+.084
)4(3)
a=100(1.02)12
a=126.82
Compound or CONTINUOUS growth:The rate is changed to a decimal if not already.
Ex: population grows continuously at a rate of 2% in Allentown. If Allentown has 10,000 people today, how many will it have 5 yearsFrom now?
A =a0ert
Compound or CONTINUOUS growth:
Ex: population grows continuously at a rate of 2% in Allentown. If Allentown has 10,000 people today, how many will it have 5 yearsFrom now?
A =a0ert
A =a0ert
A=10,000e5(.02)
A=10000e.1 ≈11052
How much will a 1000 investment be Worth in 10 years if it is compounded continuouslyat 2.5%?
If a substance decays continually at a rate of -0.05 an hour,How much will be left of a 20 gram substance in 6 hours?
How much will a 1000 investment be Worth in 10 years if it is compounded continuouslyAt 2.5%?
If a substance decays continually at a rate of -0.05 an hour,How much will be left of a 20 gram substance in 6 hours?
A =a0ert
A=1000e10(.025)
A=1284.03
A =a0ert
A=20e6(−.05)
A=14.82
Solving exponentials w/ variable exponents-like bases
• 1. decide like base
• 2. rewrite one or both numbers as a power of the base
• 3. simplify the exponents
• 4. set exponents equal and solve
• 5. be sure to check!
• Example 43x = 8x+1
Solve by equating exponents
• 43x = 8x+1
• (22)3x = (23)x+1 rewrite w/ same base
• 26x = 23x+3 power to a power: multiply
• 6x = 3x+3 set exponents equal
• x = 1 Check → 43*1 = 81+1
64 = 64
When you can’t rewrite using the same base:
1. isolate the base term2. “log” both sides
3. Solve for x
• Example: 2x = 7
• log 2x = log 7
When you can’t rewrite using the same base, you can solve by taking a log
of both sides
• 2x = 7
• log 2x = log 7
• x log 2 = log 7
• x = ≈ 2.8072log
7log
4x = 15“log” both sides & solve
4x = 15• log 4x = log 15
• x log 4 = log15
• x= log 15/log 4
• ≈ 1.95
10(2x-3) +4 = 21• Isolate the base term
10(2x-3) +4 = 21• -4 -4• 102x-3 = 17• Now solve for x….
10(2x-3) +4 = 21• -4 -4• 102x-3 = 17• log102x-3 = log17• (2x-3)(log 10) = log 17• 2x – 3 = log17/log 10• 2x =(3 +1.2304) • x= ≈ 2.115
4.2304
2
5(3)x/4 = 25• isolate and solve:
Solution:
• Ex: 5(3)x
4 =25
5(3)x
4 =25
3x4 =5
log3x4 =log5
x4
=log5log3
Solution:
• Ex: 5(3)x
4 =25
3x4 =5
log3x4 =log5
x4
=log5log3
x≈1.465(4) ≈5.860
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)
• .211 ≈ e-.046t (divide by 142)
• How do you get the variable out of the exponent?
• ln .211 ≈ ln e-.046t (take the ln of both sides)
• ln .211 ≈ -.046t ln e
• -1.556 ≈ -.046t (1)
• 33.8 ≈ t
• about 34 minutes to cool!
Cooling cont.
Domains of log equations
• Find the domain of # 19 on page 331
• Domains are important because solving logarithmic equations sometimes produces extraneous roots.
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
log3(5x-1) = log3(x+7)
Solve by decompressing
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
log5(3x + 1) = 2
• use base 5 on both sides
log5(3x + 1) = 2
• (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
log5x + log(x+1)=2• Decompress and exponentiate
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 4=x is the only solution
2
One More!
log2x + log2(x-7) = 3Solve and check:
One More!
log2x + log2(x-7) = 3• log2x(x-7) = 3• log2 (x2- 7x) = 3• 2log
2(x -7x) = 23
• x2 – 7x = 8• x2 – 7x – 8 = 0• (x-8)(x+1)=0• x=8 x= -1
2
Page 331 # 38
• Solve and ck.
• log(x-2) + log(x+5) =2log 3
Solution:
• log(x-2) + log(x+5) =2log 3
• (x-2)(x+5)=9
• x2 + 3x -19 = 0
• (the neg. answer is extraneous)
x =−3+ 85
2≈3.1
Assignment
Hw pg 331 18-25, 33-37 odds