Solving equations involving exponents and logarithms
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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
Related Documents
