6.10/6.11 Laws of Logarithms andchange of base formula

Laws of Logarithms(you don’t need to write exp. Laws)

m n m nb b b log log logb b bMN M N

Exponent Laws Logarithm Laws

mm n

n

b bb

log log logb b bM M NN

log logkb bM k M

Express as a single logarithm

10 10log 3logp q3

10 10log logp q 3

10log pq

10 104log 2logp q4 2

10 10log logp q 4

10 2log pq

Ex) TOO)

Example

2 36log M N

2 36 6 6Express log in terms of log and logM N M N

2 36 6log logM N

6 62log 3logM N

Ex )

10) log 18a

If log 2 0.30 and log3 0.48, find:

10log (2 9) 2

10log (2 3 ) 2

10 10log 2 log 3

10 10log 2 2log 3 0.30 2(0.48)

1.260.30 0.96

1020) log3

b

10 10 10log 2 log 10 log 3

102 10log

3

0.30 1 0.48 1.30 0.48 0.82

Use log1010 = 1

TOO)

log 5

If log 2 0.30 and log3 0.48, find:

1010log2

10 10log 10 log 2

1 0.30 0.70

Use log1010 = 1

Facts on Logs

log 1 0b

log 1b b

log kb b k

logb Nb NNEW !!!

Change-of-Base Formula

logloglogaxxa

base down low

use your calculatorbase 10

Ex) 104

10

log 7log 7log 4

1.403...

log button on your calculator is base 10!!

Change-of-Base Formula**Remember no CASIOS on this test!

2 7x

base down low

Write in log formGo around the world Starting with base

Ex)

2log 7 x

2.807...log 7log 2

x

TOO: Use Change-of-Base Formulato solve:

base down low

5log 8 x1.292...

log8log5

x

2 24log 3 log 9Write as a single log

42 2

2 2

log 3 log 981log log 99

Homework

Pg. 278-279 #1-49 oddPg. 284-285 #1-19 odd