1. 5.3 Properties of Logarithms
Goals
Use the base change 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. Base Change Formula To change to base 10… log x log b x = log b
To change to base e…
ln x
ln b
log b x =

3. Using the Change of Base Formula
Example:
log 25
log 4
log4 25 = =
Rewrite using the base change formula.
log 12
log 2
log2 12 = =
FYI…You can solve the two problems above using the natural logarithm as well.
ln 25
ln 4
log4 25 = =

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
Rewrite
Example
log5 35
Bring exponent out front.
log5 (5)1/3
1/3log5 (5)
1/3 ( 1)
= 1/3

6. Bring exponent out front. Use your log properties
Example #2:
ln e6 – ln e2
6ln e – 2ln e
6(1) – 2(1)
Use the division property
6 - 2
Alternative Method for Example #2 4
Bring exponent out front.
ln e6 – ln e2 ln e6 e2
ln e4
4 ln e
4 (1)
= 4

7. Using Properties of Logarithms to expand the expression as a sum, difference and/or constant2 27 ln
= ln 2 - ln 27
log310z
= log310 + log3z
log 4x2y
= log 4 + log x2 + log y
= log 4 + 2log x + log y 6
 x2 + 1 ln
= ln 6 – ln (x2 + 1)1/2
= ln 6 – 1/2ln (x2 + 1)

8. Your Turn:
Use the Properties of Logarithms to expand the expression as a sum, difference and/or constant.
log4 5x3y = log log4 x + log4 y
3x – 5 7
ln = 1/2 ln(3x – 5) – ln 7

9. Write the expression as a single logarithm (Go Backwards)
ln y + ln t
= ln yt 8 t
log 8 – log t
= log
-4ln 2xt
= ln (2xt)-4
2 ln ln (x – 4)
= ln 82 + ln (x – 4)5
= ln 82(x – 4)5
= ln 64(x – 4)5
1/3[log x + log (x + 1)]
= 1/3 [log x(x + 1)]
= log [x(x + 1)] 1/3
= log 3 x(x + 1)]

10. = ln x6 (x + 1)2 (x – 1) 2 Try this one…
2[3ln x – ln (x + 1) – ln(x – 1)]
= 2[ln x3 – ln (x + 1) – ln(x – 1)]
= 2ln x3 – 2ln (x + 1) – 2ln(x – 1)
= ln x6 – ln (x + 1)2 –ln(x – 1) 2 x6
(x + 1)2
= ln – ln(x – 1) 2
(x – 1) 2 1 x6
(x + 1)2
= ln ÷ x6
(x + 1)2 1
(x – 1) 2
= ln