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Properties of Logarithms Tools for solving logarithmic and exponential equations
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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