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.

To change to base 10…

log b x = log xlog b

To change to base e…

log b x = ln xln b

Example:

log4 25 =

Rewrite using the base change formula.

log 25 log 4

= 2.3219

log2 12 =log 12 log 2

= 3.5850

FYI…You can solve the two problems above using the natural logarithm as well. log4 25 =ln 25

ln 4= 2.3219

Product Property: loga (uv) = loga u + loga v

Quotient Property: loga (u/v) = loga u - loga v

Power Property: loga un = n loga u

log5 (5)1/3

1/3 ( 1)

Examplelog5 35

RewriteBring

exponent out front.

1/3log5 (5)

= 1/3

Alternative Method for Example #2

Example #2:ln e6 – ln e2

Bring exponent out

front.

6ln e – 2ln e

Use your log properties

6(1) – 2(1)6 - 24

ln e6 – ln e2

Use the division property ln e

6

e2

ln e4

Bring exponent out front.

4 ln e4 (1) = 4

ln = ln 2 - ln 27

log310z = log310 + log3z

ln

log 4x2y = log 4 + log x2 + log y = log 4 + 2log x + log

y = ln 6 – ln (x2 + 1)1/2

= ln 6 – 1/2ln (x2 + 1)

227

6 x2 + 1

3x – 57

Your Turn:

Use the Properties of Logarithms to expand the expression as a sum, difference and/or constant.

log4 5x3y = log4 5 + 3 log4 x + log4 y

ln = 1/2 ln(3x – 5) – ln 7

ln y + ln t = ln yt

log 8 – log t= log-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)] = 1/3 [log x(x + 1)]

8t

= ln 64(x – 4)5

= log [x(x + 1)] 1/3

= log 3 x(x + 1)]

2[3ln x – ln (x + 1) – ln(x – 1)]

= 2[ln x3 – ln (x + 1) – ln(x – 1)]

Try this one…

= 2ln x3 – 2ln (x + 1) – 2ln(x – 1)

= ln x6 – ln (x + 1)2 –ln(x – 1) 2

= ln – ln(x – 1) 2 x6

(x + 1)2

= ln ÷x6

(x + 1)2

(x – 1) 2

1

= ln x6

(x + 1)2

1(x – 1) 2

= ln x6

(x + 1)2 (x – 1) 2