Properties of logarithms
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Transcript of Properties of logarithms
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