Properties of logarithms

Math 3

Keeper 29

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

EXAMPLE 1: Use log53≈.683 and log57≈1.209 to

approximate the following:

a) log53/7 =

log53 – log57 ≈

0.683 – 1.209 =

-0.526

b) log521 = log5(3·7)= log53 + log57≈0.683 + 1.209 = 1.892

EXAMPLE 1 (continued): Use log53≈.683 and log57≈1.209 to

approximate the following:c) log549 =

log572 =

2 log57 ≈

2(1.209)=

2.418

YOUR TURN!Use log95≈0.732 and log911≈1.091

to approximate the following:

d) log95/11

e) log955

f) log925

d) -0.359 e) 1.823

f) 1.464

EXAMPLE 2: Expand the given logarithm

*You can use the properties to expand logarithms.

a) log2 =

log27x3 - log2y =

log27 + log2x3 – log2y =

log27 + 3·log2x – log2y

y

x37

Your turn!Expand the logarithm.

b) log 5mn =

log 5 + log m + log n

c) log58x3 =

log58 + 3·log5x

EXAMPLE 3: Condense the logarithms

a) log 6 + 2 log2 – log 3 =

log 6 + log 22 – log 3 =

log (6·22) – log 3 =

log =

log 8

3

26 2

YOUR TURN AGAIN! Condense the logarithm.

b) log57 + 3·log5t =

log57t3

c) 3log2x – (log24 + log2y)=

log2 y

x

4

3

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)

c

u

b

b

log

log

c

u

log

log

c

u

ln

ln

EXAMPLE 4: Evaluate using the change-of-base formula using 1) common log & 2) natural logarithm.

a1) log37 =

log 7 ≈

log 3

0.8451 ≈

0.4771

1.771

a2) ln 7 ≈ ln 3

1.946 ≈ 1.099 1.771

EXAMPLE 4 (continued): Evaluate using the change-of-base formula using 1) common log & 2) natural logarithm.

b1) log48 =

log 8 ≈

log 4

0.903 ≈

0.602

1.500

a2) ln 8 ≈ ln 4

2.079 ≈ 1.386 1.500