5.3 Properties of Logarithms Use the change of base formula to rewrite and evaluate logs Use properties of logs to evaluate or rewrite log expressions Use properties of logarithms to expand or condense logarithmic expressions Use logarithmic functions to model and solve real-life problems.

Change of Base Formula Base b Base e logax= logbx logax= ln x logba ln a

Using the Change of Base Formula Examples— log4 25 = Rewrite as: log 25 log 4 1.39794 .060206 = 2.3219 Rewrite as: log 12 log 2 1.07918 .30103 log2 12= =3.5850 The same 2 problems can be done using ln.

Properties of Logarithms Product Property: loga (uv) = loga u + loga v Quotient Property: loga (u/v) = loga u - loga v Power Property: loga un = n loga u

Using Properties of Logs to find the exact value of the expression Example log5 35 ln e6 – ln e2 Bring exponents out front. 6ln e – 2ln e Rewrite-- log5 (5)1/3 So-- 6 – 2 = 4 Bring exponent out front. 1/3 log5 (5) = 1/3 OR we could have rewritten this as division— Ln e6 = lne4 = 4lne = 4 e2

Using Properties of Logarithms to expand the expression as a sum, difference and/or constant ln 2/27 = ln 2 - ln 27 log310z = log310 + log3z log 4x2y = log 4 + log x2 + log y = log 4 + 2log x + log y ln 6 x2 + 1 = ln 6 – ln (x2 + 1)1/2 = ln 6 – 1/2ln (x2 + 1)

Write the expression as a single logarithm (Go Backwards) ln y + ln t log 8 – log t = log 8/t = ln yt -4ln 2xt = ln (2xt)-4 2 ln 8 + 5 ln (x – 4) = ln 82 + ln (x – 4)5 = ln 82(x – 4)5 1/3[log x + log (x + 1)] =[log x(x + 1)]1/3 2[3ln x – ln (x + 1) – ln(x – 1)] =[3ln x – ln (x + 1) – ln(x – 1)]2 = ln x3 2 (x + 1)(x – 1) Foil this