1. 7.5 Apply Properties of logarithms
Algebra II

2. Properties of Logarithms
Let b, u, and v be positive numbers such that b≠1.
Product property:
logbuv = logbu + logbv
Quotient property:
logbu/v = logbu – logbv
Power property:
logbun = n logbu

3. Ex. 4: log37 = log 7 ≈ log 3 1.771 ln 7 ≈ ln 3 1.771 (base 10)
Use the change of base to evaluate:
log37 =
(base 10)
log 7 ≈
log 3
1.771
(base e)
ln 7 ≈
ln 3
1.771

4. Ex. 1Use log53≈.683 and log57≈1.209 log53/7 = log521 = log53 – log57 ≈
A.)Approximate:
log53/7 =
log53 – log57 ≈
.683 – =
-.526
B.)Approximate:
log521 =
log5(3·7)=
log53 + log57≈ =
1.892

5. Use log53≈.683 and log57≈1.209
C.) Approximate:
log549 =
log572 =
2 log57 ≈
2(1.209)=
2.418

6. Ex. 2a) Expanding Logarithms
You can use the properties to expand logarithms.
log =
log27x3 - log2y =
log27 + log2x3 – log2y =
log27 + 3·log2x – log2y

7. log 5mn = log 5 + log m + log n log58x3 = log58 + 3·log5x Ex. 2
2b.) Expand:
log 5mn =
log 5 + log m + log n
2c.) Expand:
log58x3 =
log58 + 3·log5x

8. Ex. 3a.) Condensing Logarithms
log log2 – log 3 =
log 6 + log 22 – log 3 =
log (6·22) – log 3 =
log =
log 8

9. Ex. 3 continued log57 + 3·log5t = log57t3 3log2x – (log24 + log2y)=
3b.) Condense:
log57 + 3·log5t =
log57t3
3c.) Condense:
3log2x – (log24 + log2y)=
log2

10. Change of base formula:
u, b, and c are positive numbers with b≠1 and c≠1. Then:
logcu =
logcu = (base 10)
logcu = (base e)

11. Assignment