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Properties of Logarithms Tools for solving logarithmic and exponential equations Let’s review some terms. When we write log5 125 5 is called the base 125 is called the argument 2 Logarithmic form of 5 = 25 is log525 = 2 For all the laws a, 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 : loga1 = 0 0 since a = 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 1 since a = a log 55 = ? log 55 = ? logaa = 1 log 55= 1 logaa = 1 ln e = ? ln e = logee = ? ln means loge ln e = logee = ? logaa = 1 ln e = 1 logaa = 1 Just a tiny bit harder : r logaa = r r r since a = a ln 3x = e ? ln 3x = e loge 3x e =? ln means loge ln 3x = e loge 3x e =? log a r r a ln 3x = e loge 3x e = 3x log a r r a log(105y) = ? log(105y) = ? log means log10 log(105y) = log10 105y = ? log means log10 log(105y) = log10 105y = ? log a r r a log(105y) = log10 105y = ? log a r r a log(105y) = log10 105y = 5y log a r r a log a MN log a M log a N Evidence that it works (not a proof): log log 125 3 125 log 5 25 log 5 5 5 5 3 2 1 log M log a N M a N a Evidence that it works (not a proof): log log log 2 25 2 125 5 log 5 125 log 5 5 5 5 3 1 log(2x) = ? log(2x) = ? log a MN log a M log a N log(2x) = log(2) + log(x) log a MN log a M log a N 2 ln ? x 3 2 ln ? x 3 M log a log N a M log a N 2 ln ln 2 ln x 3 x 3 M log a log N a M log a N Power Rule : r logaM = r logaM Think of it as repeated uses log MM log M log M 2 log ( M ) of r times a a a a ln( x ) ? 2 ln( x ) ? 2 log M r r a log a M 2 ln( x) ln x 2 log M r r a log a M ln x y ? 2 log MN log M log N ln x y ? 2 log MN log M log N ln x y ln( x ) ln ( y) 2 2 log MN log M log N ln x y ln( x ) ln ( y) 2 2 log M r r a log a M ln x y ln( x ) ln ( y) 2 2 2 ln( x) ln ( y ) log M r r a log a M 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) log MN log M log N 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 that logaM = 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 = 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 2x 3 = x 5 2x 3 = x 5 If M = then N ln M = ln N 2x 3 2x ln(3 ) = x 5 = x ln(5 ) If M = then N ln M = ln N 2x 3 2x ln(3 ) = x 5 = x ln(5 ) log a M r r log a M 2x 3 x 5 = ln(32x) = ln(5x ) 2x ln(3 ) = x ln(5) log a M r r log a M 2x 3 x 5 = ln(32x) = ln(5x ) 2x ln(3 ) = x ln(5) simple algebra 2x 3 x 5 = ln(32x) = ln(5x ) 2x ln(3 ) = x ln(5) 2x(ln 3) – x ln(5) = 0 simple algebra 2x 3 x 5 = ln(32x) = ln(5x ) 2x ln(3 ) = x ln(5) 2x(ln 3) – x ln(5) = 0 x[2ln(3) – ln(5)] = 0 factor out x 2x 3 x 5 = ln(32x) = ln(5x ) 2x ln(3 ) = x ln(5) 2x(ln 3) – x ln(5) = 0 x[2ln(3) – ln(5)] = 0 x 0 2 ln( 3) ln( 5) Divide out numerical coefficient 2x 3 x 5 = ln(32x) = ln(5x ) 2x ln(3 ) = x ln(5) 2x(ln 3) – x ln(5) = 0 x[2ln(3) – ln(5)] = 0 x 0 2 ln( 3) ln( 5) =0 Simplify the fraction log a M ln M ln a Change of Base Formula : When you need to approximate log53 log a M ln M ln a Change of Base Formula : When you need to approximate log53 log 5 3 ln 3 ln 5 Here’s one not seen as much as some of the others: a loga M M Here’s an example a log a M e M ln 3 x 3x