Laws of logarithms
Expanding log expressions
Combining log expressions
Quiz Review
Chapter 4.4: Laws of logarithms
Every law of exponents gives rise to a corresponding law of logarithms. In the following,
suppose a > 0 and a 6= 1, and let a be the base for the logarithm function loga (x).
Also assume that u and v are > 0, so that loga u and loga v are defined.
Laws of logarithms
The function y = ax has an inverse function y = loga (x), defined for x > 0.
• Basic principle of logarithms: loga ax = x.
To find loga x, express x as a power of a.
•
•
•
•
•
loga (1) = loga (a0 ) = 0 and loga (a) = loga (a1 ) = 1.
loga uv = loga u + loga v: Log of product = sum of logs.
loga uv = loga u − log v: Log of quotient = difference of logs.
loga v1 = − loga v: Log of reciprocal = negative of log.
Let n be any real number. Then loga un = n loga u:
To find the log of a power, bring down the power in front.
Stanley Ocken
M19500 Precalculus Chapter 4.4 Laws of logarithms
Laws of logarithms
Expanding log expressions
Combining log expressions
Quiz Review
The inverse of y = 2x is y = log2 (x). Its defining property is that log2 (2x ) = x.
√
Example 1: Find log2 (16), log2 ( 18 ), and log2 (8 2)
Solutions: • log2 (16) = log2 (24 ) = 4.
• 18 = 213 = (2−3 ), and so log2 ( 18 ) = log2 (2−3 ) = −3.
√
√
1
7
7
• 8 2 = 23 · 21/2 = 23+ 2 = 2 2 , and so log2 (8 2) = log2 (2 2 ) = 72 .
The inverse of y = 10x is y = log10 (x). Its defining property is that log10 (10x ) = x.
Example 2: Find log10 (1000000) and log10 (.0001)
Solutions: • log10 (1000000) = log10 (106 ) = 6.
• log10 (.0001) = log10 ( 1014 ) = log10 (10−4 ) = −4.
The inverse of y = ex is y = ln(x). Its defining property is that ln(ex ) = x.
Example 3: Find ln(1), ln(e), and ln( √1e ).
Solutions: • ln(1) = ln(e0 ) = 0.
• ln(e) = ln(e1) =1.
1
• ln( √1e ) = ln 11 = ln(e− 2 ) = − 21 .
e2
Stanley Ocken
M19500 Precalculus Chapter 4.4 Laws of logarithms
Laws of logarithms
Expanding log expressions
Combining log expressions
Quiz Review
h
√
√ i
Example 4: Use log uv = log u + log v to find log (x2 + 1)4 x2 + 1 3 x
h
√
√ i
Solution: Original problem:
log (x2 + 1)4 x2 + 1 3 x
1
1
Apply log uv = log u + log v:
Bring down the powers:
= log (x2 + 1)4 + log (x2 + 1) 2 + log x 3
= 4 log(x2 + 1) + 21 log(x2 + 1) + 13 log x
Combine like terms:
=
9
2
log(x2 + 1) + 13 log x .
Example 5: Use log uv = log u − log v to find log √
Solution: Original problem:
Apply log uv = log u − log v
Apply log uv = log u + log v
Make sure to use parentheses!
Bring down the powers:
Distribute minus sign
x3
x2 + 1(x + 2)3
3
log √x2 +1x (x+2)3
√
=log x3 − log( x2 + 1 (x + 2)3 )
1
= log x3 − ( log (x2 + 1) 2 + log(x + 2)3 )
= 3 log x − 12 log(x2 + 1) + 3 log(x + 2)
= 3 log x − 21 log(x2 + 1) − 3 log(x + 2) .
Stanley Ocken
M19500 Precalculus Chapter 4.4 Laws of logarithms
Laws of logarithms
Expanding log expressions
Combining log expressions
√
3
Example 6: Use laws of logarithms to expand and simplify log a5 bc3
Solution:
√
3
The expression to expand is:
log a5 bc3
1
Rewrite the cube root as the power 31 = log (a5 bc3 ) 3
Apply log un = n log u
= 31 log(a5 bc3 )
Apply log uvw = log u + log v + log w = 31 (log a5 + log b + log c3 )
Apply log un = n log u
= 31 (5 log a + log b + 3 log c)
Multiply out
=
5
3
log a + 13 log b + log c
3
√
Example 7: Use laws of logarithms to expand and simplify log abc
3
b
Solution:
The expression to expand is:
Apply
Apply
Apply
Apply
3
√
log abc
3
b
log uv = log u − log v:
log un = n log u
log uvw = log u + log v + log w
log u3 = 3 log u
Note:
√
1
3
b = b3
1
= log abc3 − log b 3
= log abc3 − 31 log b
= log a + log b + log c3 − 13 log b
= log a + log b + 3 log c − 31 log b
= log a + 3 log c + 23 log b
Combine terms
Stanley Ocken
M19500 Precalculus Chapter 4.4 Laws of logarithms
Quiz Review
Laws of logarithms
Expanding log expressions
Combining log expressions
Quiz Review
In the previous examples, we expanded the log of a complicated expression as the sum
or difference of logs of simpler expressions. The following examples illustrate the
reverse process:
Example 8: Find log8 2 + log8 32.
Easy solution: Combine terms by using the logarithm product formula:
log8 u + log8 v = log8 uv with u = 2 and u = 32.
log8 2 + log8 32 = log8 2 · 32 = log8 64 = log8 82 = 2 .
Hard solution:
• Since 23 = 8, then 2 = 81/3 and so log8 2 = log8 (81/3 ) = 31 .
1
5
• Since 32 = 25 , then 32 = (8 3 )5 = 8 3 and so log8 (32) = log8 (85/3 ) = 35 .
Therefore log8 2 + log8 32 = 31 + 53 = 63 = 2 .
Stanley Ocken
M19500 Precalculus Chapter 4.4 Laws of logarithms
Laws of logarithms
Expanding log expressions
Combining log expressions
In the following two examples, use: sum of logs = log of product.
Example 9: Use log laws to combine log a + log b into a single expression.
Solution: log a + log b = log ab
Example 10: Use log laws to combine log a + 12 log b into a single expression.
√
1
1
Solution: log a + 21 log b = log a + log b 2 = log ab 2 = log a b
In the following two examples, use: difference of logs = log of quotient.
Example 11: Use log laws to combine log a − log b into a single expression.
Solution: log a − log b = log ab
Example 12: Use log laws to combine log a − log b − log c into a single expression.
a
Solution: log a − log b − log c = log a − (log b + log c) = log a − log bc = log bc
Stanley Ocken
M19500 Precalculus Chapter 4.4 Laws of logarithms
Quiz Review
Laws of logarithms
Expanding log expressions
Combining log expressions
Quiz Review
Example 13: Use log laws to combine 3 log a + 2 log b + 4 log c into a single expression.
Solution:
The expression to combine is:
Apply n log u = log un :
3 log a + 2 log b + 4 log c
= log a3 + log b2 + log c4
Apply log u + log v + log w = log uvw :
= log a3 b2 c4
Example 14: Use laws of logarithms to combine 3 ln x + 31 ln y − 4 ln(t2 + 1) into a
single expression.
Solution:
The expression to expand is:
Apply n log u = log un :
Apply log u + log v = log uv
Apply log u − log v = log uv
3 ln x + 13 ln y − 4 ln(t2 + 1)
1
= ln x3 + ln y 3 − ln(t2 + 1)4
1
= ln x3 y 3 − ln(t2 + 1)4
!
1
x3 y 3
= ln
(t2 + 1)4
Stanley Ocken
M19500 Precalculus Chapter 4.4 Laws of logarithms
Laws of logarithms
Expanding log expressions
Combining log expressions
Quiz Review
√
Example 1: Find log2 (16), log2 ( 18 ), and log2 (8 2).
Example 2: Find log10 (1000000) and log10 (.0001).
Example 3: Find ln(1), ln(e), and ln( √1e ).
3
√
.
Example 4: Use laws of logarithms to expand and simplify log abc
3
b
h
√
√ i
Example 5: Use log uv = log u + log v to find log (x2 + 1)4 x2 + 1 3 x .
3
x
Example 6: Use log uv = log u − log v to find log √x2 +1(x+2)
.
3
Example 7: Use laws of logarithms to expand and simplify log
Stanley Ocken
√
3
a5 bc3 .
M19500 Precalculus Chapter 4.4 Laws of logarithms
Quiz Review
Laws of logarithms
Expanding log expressions
Combining log expressions
Quiz Review
Example 8: Find log8 2 + log8 32.
Example 9: Use log laws to combine log a + log b into a single expression.
Example 10: Use log laws to combine log a + 12 log b into a single expression.
Example 11: Use log laws to combine log a − log b into a single expression.
Example 12: Use log laws to combine log a − log b − log c into a single expression.
Example 13: Use log laws to combine 3 log a + 2 log b + 4 log c into a single expression.
Example 14: Use laws of logarithms to combine 3 ln x + 31 ln y − 4 ln(t2 + 1) into a
single expression.
Stanley Ocken
M19500 Precalculus Chapter 4.4 Laws of logarithms