# Properties of Logarithms

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07-Feb-2016Category

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### Transcript of Properties of Logarithms

Properties of LogarithmsTools for solving logarithmic and exponential equations

Lets review some terms.

When we write log5 125

5 is called the base125 is called the argument

Logarithmic form of 52 = 25 is

log525 = 2

For all the laws a, M and N > 0a 1r is any real

Remember ln and logln is a short cut for logelog means log10

Easy ones first : loga1 = 0 since a0 = 1

log 31= ?

log 31= ? loga1 = 0

log 31= 0 loga1 = 0

ln 1 = ?

ln 1 = ? loga1 = 0

ln 1 = 0 loga1 = 0

Another easy one : logaa = 1 since a1 = a

log 55 = ?

log 55 = ?logaa = 1

log 55= 1logaa = 1

ln e = ?

ln e = logee = ? ln means loge

ln e = logee = ?logaa = 1

ln e = 1logaa = 1

Just a tiny bit harder : logaar = r since ar = ar

ln e3x = ?

ln e3x = loge e3x = ? ln means loge

ln e3x = loge e3x = ?

ln e3x = loge e3x = 3x

log(105y) = ?

log(105y) = ? log means log10

log(105y) = log10 105y = ? log means log10

log(105y) = log10 105y = ?

log(105y) = log10 105y = ?

log(105y) = log10 105y = 5y

Evidence that it works (not a proof):

Evidence that it works (not a proof):

log(2x) = ?

log(2x) = ?

log(2x) = log(2) + log(x)

Power Rule : logaMr = r logaM

Think of it as repeated uses of r times

NEVER DO THISlog ( 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)

Consider 5 = 5

You know that the

and the are equal

So if you knew that : logaM = logaN you would know that M = N

And vice versa, suppose M = N Then it follows thatlogaM = logaN

ln (x + 7) = ln(10)

ln (x + 7) = ln(10)x+7 = 10 ln(M) = ln (N)

ln (x + 7) = ln(10)x+7 = 10x = 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 - 49 = x oh, this step is easy

32x = 5x

If M = N then ln M = ln N32x = 5x

If M = N then ln M = ln N32x = 5xln(32x) = ln(5x )

32x = 5xln(32x) = ln(5x )

32x = 5xln(32x) = ln(5x )2x ln(3 ) = x ln(5)

simple algebra 32x = 5xln(32x) = ln(5x )2x ln(3 ) = x ln(5)

simple algebra 32x = 5xln(32x) = ln(5x )2x ln(3 ) = x ln(5)2x(ln 3) x ln(5) = 0

factor out x 32x = 5xln(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 = 5xln(32x) = ln(5x )2x ln(3 ) = x ln(5)2x(ln 3) x ln(5) = 0x[2ln(3) ln(5)] = 0

Simplify the fraction 32x = 5xln(32x) = ln(5x )2x ln(3 ) = x ln(5)2x(ln 3) x ln(5) = 0x[2ln(3) ln(5)] = 0=0

Change of Base Formula : When you need to approximate log53

Change of Base Formula : When you need to approximate log53

Heres one not seen as much as some of the others:

Heres an example

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