Solving equations involving exponents and logarithms.
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Transcript of Solving equations involving exponents and logarithms.
![Page 1: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/1.jpg)
Solving equations involving exponents and logarithms
![Page 2: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/2.jpg)
Let’s review some terms.
When we write log
5 125
5 is called the base125 is called the argument
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Logarithmic form of 52 = 25 is
log525 = 2
![Page 4: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/4.jpg)
For all the lawsa, M and N > 0
a ≠ 1
r is any real
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Remember ln and log
ln is a short cut for loge
log means log10
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Log laws
a
MM
NMN
M
NMMN
MrM
a
a
aaa
aaa
ar
a
a
a
ln
lnlog
logloglog
logloglog
loglog
1log
01log
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If your variable is in an exponent or in the argument of a logarithm
Find the pattern your equation resembles
NMNM
ennb be
lnln
log
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If your variable is in an exponent or in the argument of a logarithm
Find the pattern
Fit your equation to match the pattern Switch to the equivalent form Solve the result Check (be sure you remember the domain of a log)
NMNM
ennb be
lnln
log
![Page 9: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/9.jpg)
2
)5ln(x
log(2x) = 3
![Page 10: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/10.jpg)
It fits
2
)5ln(x
log(2x) = 3
enb log
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Switch
2
)5ln(x
log(2x) = 3
103=2x ennb be log
Did you remember that log(2x) means log10(2x)?
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Divide by 2
2
)5ln(x
log(2x) = 3
103=2x
500 = x
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ln(x+3) = ln(-7x)
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ln(x+3) = ln(-7x)
It fitsNM lnln
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ln(x+3) = ln(-7x)
Switch
NMNM lnln
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ln(x+3) = ln(-7x)
x + 3 = -7x
Switch
NMNM lnln
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ln(x+3) = ln(-7x)
x + 3 = -7x
x = - ⅜
Solve the result
(and check)
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ln(x) + ln(3) = ln(12)
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ln(x) + ln(3) = ln(12)
x + 3 = 12
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ln(x) + ln(3) = ln(12)
x + 3 = 12
Oh NO!!! That’s wrong!
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You need to use log laws
ln(x) + ln(3) = ln(12)
ln(3x) = ln (12)
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Switch
ln(x) + ln(3) = ln(12)
ln(3x) = ln (12)
3x = 12
NMNM lnln
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ln(x) + ln(3) = ln(12)
ln(3x) = ln (12)
3x = 12
x = 4 Solve the result
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log3(x+2) + 4 = 9
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It will fit
log3(x+2) + 4 = 9
enb log
![Page 26: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/26.jpg)
Subtract 4 to make it fit
log3(x+2) + 4 = 9
log3(x+2) = 5
enb log
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Switch
log3(x+2) + 4 = 9
log3(x+2) = 5
nben eb log
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Switch
log3(x+2) + 4 = 9
log3(x+2) = 5
35 = x + 2
nben eb log
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Solve the result
log3(x+2) + 4 = 9
log3(x+2) = 5
35 = x + 2
x = 241
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5(10x) = 19.45
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Divide by 5 to fit
5(10x) = 19.45
10x = 3.91
nbe
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Switch
5(10x) = 19.45
10x = 3.91
ennb be log
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Switch
5(10x) = 19.45
10x = 3.91
log(3.91) = x
ennb be log
![Page 34: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/34.jpg)
Exact log(3.91)
Approx 0.592
5(10x) = 19.45
10x = 3.91
log(3.91) = x
≈ 0.592
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2 log3(x) = 8
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It will fit
2 log3(x) = 8
enb log
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Divide by 2 to
fit
2 log3(x) = 8
log3(x) = 4
enb log
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Switch
2 log3(x) = 8
log3(x) = 4
nben e
b log
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Switch
2 log3(x) = 8
log3(x) = 4
34=x
nben eb log
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Then Simplify
2 log3(x) = 8
log3(x) = 4
34=x
x = 81
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log2(x-1) + log2(x-1) = 3
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Need to use a log law
log2(x-1) + log2(x-1) = 3
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log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
NMMN logloglog
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Switch
log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
nben e
b log
![Page 45: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/45.jpg)
Switch
log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
23=(x-1)(x+1)
nben e
b log
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and finish
log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
23=(x-1)(x+1) = x2 -1
x = +3 or -3
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But -3 does not check!
log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
23=(x-1)(x+1) = x2 -1
x = +3 or -3
![Page 48: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/48.jpg)
Exclude -3 (it would
cause you to have a negative argument)
log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
23=(x-1)(x+1) = x2 -1
x = +3 or -3
![Page 49: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/49.jpg)
There’s more than one way to do this
13ln x
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389.4
3
3
2)3ln(
1)3ln(2
1
1)3ln(
13ln
2
2
2
1
x
xe
xe
x
x
x
x
Can you find why each step is valid?
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389.4
3
3
2)3ln(
1)3ln(2
1
1)3ln(
13ln
2
2
2
1
x
xe
xe
x
x
x
x
rules of exponents
multiply both sides by 2
- 3 to get exact answer
Approximate answer
MrM ar
a loglog
ennb be log
![Page 52: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/52.jpg)
Here’s another way to solve the same equation.
389.4
3
3
3
3
13ln
2
2
22
1
x
xe
xe
xe
xe
x
![Page 53: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/53.jpg)
exclude 2nd result
389.43
33
33
3
3
3
13ln
2
22
22
2
22
1
ex
xeorxe
xeorxe
xe
xe
xe
x
Square both sides
Simplify
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52x - 5x – 12 = 0
![Page 55: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/55.jpg)
Factor it. Think of
y2 - y-12=0
52x - 5x – 12 = 0
(5x – 4)(5x + 3) = 0
![Page 56: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/56.jpg)
Set each factor = 0
52x - 5x – 12 = 0
(5x – 4)(5x + 3) = 0
5x – 4 = 0 or 5x + 3 = 0
![Page 57: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/57.jpg)
Solve first factor’s equation
Solve 5x – 4 = 0
5x = 4
log54 = x
![Page 58: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/58.jpg)
Solve other factor’s
equation
Solve 5x + 3 = 0
5x = -3
log5(-3) = x
![Page 59: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/59.jpg)
Oops, we cannot have a
negative argument
Solve 5x + 3 = 0
5x = -3
log5(-3) = x
![Page 60: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/60.jpg)
Exclude this solution.
Only the other factor’s solution
works
Solve 5x + 3 = 0
5x = -3
log5(-3) = x
![Page 61: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/61.jpg)
2
)5ln(x
4x+2 = 5x
![Page 62: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/62.jpg)
If M = N then
ln M = ln N
2
)5ln(x
4x+2 = 5x
![Page 63: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/63.jpg)
2
)5ln(x
If M = N then ln M = ln N
4x+2 = 5x
ln(4x+2) = ln(5x )
![Page 64: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/64.jpg)
4x+2 = 5x
ln(4x+2) = ln(5x)
MrM ar
a loglog
![Page 65: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/65.jpg)
4x+2 = 5x
ln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )
MrM ar
a loglog
![Page 66: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/66.jpg)
Distribute
4x+2 = 5x
ln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )
x ln (4) + 2 ln(4) = x ln (5)
![Page 67: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/67.jpg)
Get x terms on one side
4x+2 = 5x
ln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )
x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4)
![Page 68: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/68.jpg)
Factor out x
4x+2 = 5x
ln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )
x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4)
![Page 69: Solving equations involving exponents and logarithms.](https://reader035.fdocuments.in/reader035/viewer/2022062314/56649d5e5503460f94a3d51a/html5/thumbnails/69.jpg)
Divide by numerical coefficient
4x+2 = 5x
ln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )
x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4)
)5ln()4ln(
)4ln(2
x