1. 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.

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

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

4. 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

5. 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

6. 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
x2 + 1
= ln 6 – ln (x2 + 1)1/2
= ln 6 – 1/2ln (x2 + 1)

7. 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 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 x
(x + 1)(x – 1)
Foil this