http://www.ck12.org Chapter 3. Logs and Exponents
logab=log^1 ba
Solution:
logab=loglogxxba= loglog^1 xa
xb
=log^1 baExample C
Simplify to an exact result:(log 45 )·(log 34 )·(log 581 )·(log 525 )
Solution:
log 5log 4·log 4log 3·log 3log 5^4 ·log 5log 5^2 =log 5log 4·log 4log 3· 4 ·log 5log 3· (^2) log 5·log 5 4 · 2 = 8
Concept Problem Revisited
In order to evaluate an expression like log 3 12 you have two options on your calculator:
ln 12ln 3 =log 12log 3 ≈ 2. 26
Vocabulary
Change of baserefers to the formula that allows you to rewrite a logarithm with a new base that you choose. Com-
mon bases to use are 10 ande.
Guided Practice
- Simplify the following expression:(log 20137 +log 20139 −log 201311 )log 2013
- Evaluate: log 248 −log 436
- Given log 35 ≈ 1 .465 find log 25 27 without using a log button on the calculator.
Answers:
[ log 7
log 2013+log 2013log 9 −log 2013log 11]
·log 2013=log 7+log 9−log 11=log(^711 ·^9 )- log 248 −log 436
=log 48log 2−log 36log 2 2=log 48log 2−log 62
log 2^2
=log 48log 2−^22 ··log 6log 2=log 48log 2−log 6=log( 48
6)
log 2
=log 8log 2=log 23
log 2
=^3 log 2·log 2
= 3