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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