solving equations involving exponents and logarithms
Post on 15-Jan-2016
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Solving equations involving exponents and logarithms
Let’s review some terms.
When we write log
5 125
5 is called the base125 is called the argument
Logarithmic form of 52 = 25 is
log525 = 2
For all the lawsa, M and N > 0
a ≠ 1
r is any real
Remember ln and log
ln is a short cut for loge
log means log10
Log laws
a
MM
NMN
M
NMMN
MrM
a
a
aaa
aaa
ar
a
a
a
ln
lnlog
logloglog
logloglog
loglog
1log
01log
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
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
2
)5ln(x
log(2x) = 3
It fits
2
)5ln(x
log(2x) = 3
enb log
Switch
2
)5ln(x
log(2x) = 3
103=2x ennb be log
Did you remember that log(2x) means log10(2x)?
Divide by 2
2
)5ln(x
log(2x) = 3
103=2x
500 = x
ln(x+3) = ln(-7x)
ln(x+3) = ln(-7x)
It fitsNM lnln
ln(x+3) = ln(-7x)
Switch
NMNM lnln
ln(x+3) = ln(-7x)
x + 3 = -7x
Switch
NMNM lnln
ln(x+3) = ln(-7x)
x + 3 = -7x
x = - ⅜
Solve the result
(and check)
ln(x) + ln(3) = ln(12)
ln(x) + ln(3) = ln(12)
x + 3 = 12
ln(x) + ln(3) = ln(12)
x + 3 = 12
Oh NO!!! That’s wrong!
You need to use log laws
ln(x) + ln(3) = ln(12)
ln(3x) = ln (12)
Switch
ln(x) + ln(3) = ln(12)
ln(3x) = ln (12)
3x = 12
NMNM lnln
ln(x) + ln(3) = ln(12)
ln(3x) = ln (12)
3x = 12
x = 4 Solve the result
log3(x+2) + 4 = 9
It will fit
log3(x+2) + 4 = 9
enb log
Subtract 4 to make it fit
log3(x+2) + 4 = 9
log3(x+2) = 5
enb log
Switch
log3(x+2) + 4 = 9
log3(x+2) = 5
nben eb log
Switch
log3(x+2) + 4 = 9
log3(x+2) = 5
35 = x + 2
nben eb log
Solve the result
log3(x+2) + 4 = 9
log3(x+2) = 5
35 = x + 2
x = 241
5(10x) = 19.45
Divide by 5 to fit
5(10x) = 19.45
10x = 3.91
nbe
Switch
5(10x) = 19.45
10x = 3.91
ennb be log
Switch
5(10x) = 19.45
10x = 3.91
log(3.91) = x
ennb be log
Exact log(3.91)
Approx 0.592
5(10x) = 19.45
10x = 3.91
log(3.91) = x
≈ 0.592
2 log3(x) = 8
It will fit
2 log3(x) = 8
enb log
Divide by 2 to
fit
2 log3(x) = 8
log3(x) = 4
enb log
Switch
2 log3(x) = 8
log3(x) = 4
nben e
b log
Switch
2 log3(x) = 8
log3(x) = 4
34=x
nben eb log
Then Simplify
2 log3(x) = 8
log3(x) = 4
34=x
x = 81
log2(x-1) + log2(x-1) = 3
Need to use a log law
log2(x-1) + log2(x-1) = 3
log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
NMMN logloglog
Switch
log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
nben e
b log
Switch
log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
23=(x-1)(x+1)
nben e
b log
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
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
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
There’s more than one way to do this
13ln x
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?
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
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
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
52x - 5x – 12 = 0
Factor it. Think of
y2 - y-12=0
52x - 5x – 12 = 0
(5x – 4)(5x + 3) = 0
Set each factor = 0
52x - 5x – 12 = 0
(5x – 4)(5x + 3) = 0
5x – 4 = 0 or 5x + 3 = 0
Solve first factor’s equation
Solve 5x – 4 = 0
5x = 4
log54 = x
Solve other factor’s
equation
Solve 5x + 3 = 0
5x = -3
log5(-3) = x
Oops, we cannot have a
negative argument
Solve 5x + 3 = 0
5x = -3
log5(-3) = x
Exclude this solution.
Only the other factor’s solution
works
Solve 5x + 3 = 0
5x = -3
log5(-3) = x
2
)5ln(x
4x+2 = 5x
If M = N then
ln M = ln N
2
)5ln(x
4x+2 = 5x
2
)5ln(x
If M = N then ln M = ln N
4x+2 = 5x
ln(4x+2) = ln(5x )
4x+2 = 5x
ln(4x+2) = ln(5x)
MrM ar
a loglog
4x+2 = 5x
ln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )
MrM ar
a loglog
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)
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)
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)
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