Properties of Logarithms

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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 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