Properties of Logarithms
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Properties of Logarithms Tools for solving logarithmic and exponential equations
description
Properties of Logarithms. Tools for solving logarithmic and exponential equations. Let’s review some terms. When we write log 5 125 5 is called the base 125 is called the argument. Logarithmic form of 5 2 = 25 is log 5 25 = 2. For all the laws a , M and N > 0 a ≠ 1 r is any real. - PowerPoint PPT Presentation
Transcript of Properties of Logarithms
Properties of Logarithms
Tools for solving logarithmic and exponential equations
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
Easy ones first : logaa1 = 0
since a0 = 1
log
31= ?
log
31= ?
logaa1 = 0
log
31= 0
logaa1 = 0
ln 1 = ?
ln 1 = ?
logaa1 = 0
ln 1 = 0
logaa1 = 0
Another easy one : logaaa = 1
since a1 = a
log
55 = ?
log
55 = ?
logaaa = 1
log
55= 1
logaaa = 1
ln e = ?
ln e = logee = ?
ln means loge
ln e = logee = ?
logaaa = 1
ln e = 1
logaaa = 1
Just a tiny bit harder : logaaa
r = r since ar = ar
ln e3x = ?
ln e3x = loge e3x = ?
ln means loge
ln e3x = loge e3x = ?
ra ra log
ln e3x = loge e3x = 3x
ra ra log
log(105y) = ?
log(105y) = ?
log means log10
log(105y) = log10 105y = ?
log means log10
log(105y) = log10 105y = ?
ra ra log
log(105y) = log10 105y = ?
ra ra log
log(105y) = log10 105y = 5y
ra ra log
123
5log25log125log
3125log
555
5
Evidence that it works (not a proof):
NMMN aaa logloglog
NM aaNM
a logloglog
132
5log125loglog
225log
555125
5
5
Evidence that it works (not a proof):
log(2x) = ?
log(2x) = ?
NMMN aaa logloglog
log(2x) = log(2) + log(x)
NMMN aaa logloglog
?3
2ln
x
NMN
Maaa logloglog
?3
2ln
x
NMN
Maaa logloglog
3ln2ln3
2ln
x
x
Power Rule : logaaM
r = r logaaM
Think of it as repeated uses of r times
)(log2logloglog MMMMM aaaa
?)ln( 2 x
?)ln( 2 x
MrM ar
a loglog
MrM ar
a loglog
)ln(2ln 2 xx
NMMN logloglog
?ln 2 yx
NMMN logloglog
?ln 2 yx
)(ln)ln(ln 22 yxyx
NMMN logloglog
)(ln)ln(ln 22 yxyx
MrM ar
a loglog
)(ln)ln(ln 22 yxyx
MrM ar
a loglog
)(ln)ln(2 yx
NEVER DO THIS
log ( x + y) = log(x) + log(y) (ERROR)
WHY is that wrong? Log laws tell use that
log(x) + log(y) = log ( xy)Not log(x + y)
NMMN logloglog
Consider 5 = 5
You know that the
and the are equal
So if you knew that : logaaM = logaaN
you would know that
M = N
And vice versa, suppose M = N
Then it follows that
logaaM = logaaN
ln (x + 7) = ln(10)
ln (x + 7) = ln(10)
x+7 = 10
ln(M) = ln (N)
ln (x + 7) = ln(10)
x+7 = 10
x = 3 subtract 7
log3(x + 5) = log3(2x - 4)
log3(x + 5) = log3(2x - 4)
log(M) = log(N)
log3(x + 5) = log3(2x - 4)
x+5 = 2x - 4
log(M) = log(N)
log3(x + 5) = log3(2x - 4)
x+5 = 2x - 4
9 = x oh, this step is easy
32x = 5x
If M = N then ln M = ln N
32x = 5x
If M = N then ln M = ln N
32x = 5x
ln(32x) = ln(5x )
32x = 5x
ln(32x) = ln(5x )
MrM ar
a loglog
32x = 5x
ln(32x) = ln(5x )2x ln(3 ) = x ln(5)
MrM ar
a loglog
simple algebra
32x = 5x
ln(32x) = ln(5x )2x ln(3 ) = x ln(5)
simple algebra
32x = 5x
ln(32x) = ln(5x )2x ln(3 ) = x ln(5)
2x(ln 3) – x ln(5) = 0
factor out x
32x = 5x
ln(32x) = ln(5x )2x ln(3 ) = x ln(5)
2x(ln 3) – x ln(5) = 0x[2ln(3) – ln(5)] = 0
Divide out numerical coefficient
32x = 5x
ln(32x) = ln(5x )2x ln(3 ) = x ln(5)
2x(ln 3) – x ln(5) = 0x[2ln(3) – ln(5)] = 0
)5ln()3ln(2
0
x
Simplify the fraction
32x = 5x
ln(32x) = ln(5x )2x ln(3 ) = x ln(5)
2x(ln 3) – x ln(5) = 0x[2ln(3) – ln(5)] = 0
)5ln()3ln(2
0
x =0
Change of Base Formula :
When you need to approximate log53
aM
Ma ln
lnlog
aM
Ma ln
lnlog
Change of Base Formula :
When you need to approximate log53
5ln
3ln3log5
Here’s one not seen as much as some of the others:
Ma Ma log
Here’s an example
Ma Ma log
xe x 33ln
