Reading: Logarithmic Properties
In the previous section, we derived two important properties of logarithms, which allowed us to solve some basic exponential and logarithmic equations.
Exponential Property:
Change of Base:
While these properties allow us to solve a large number of problems, they are not sufficient to solve all problems involving exponential and logarithmic equations.
Difference of Logs Property:
It's just as important to know what properties logarithms do
not satisfy as to memorize the valid properties listed above. In particular, the logarithm is not a linear function, which means that it does not distribute: log(A + B) ≠ log(A) + log(B).
To help in this process we offer a proof to help solidify our new rules and show how they follow from properties you've already seen.
Let
a = logb(A) and c = logb(C), so by definition of the logarithm, ba = A and bc = C
Using these expressions,
AC = babc
Using exponent rules on the right,
AC = ba+c
Taking the log of both sides, and utilizing the inverse property of logs, log
b(AC) logb(ba+c) = a + c
Replacing
a and c with their definition establishes the result logb(AC) = logbA + logbC
The proof for the difference property is very similar.
With these properties, we can rewrite expressions involving multiple logs as a single log, or break an expression involving a single log into expressions involving multiple logs.
as a sum or difference of logs
First, noticing we have a quotient of two expressions, we can utilize the difference property of logs to write
Then seeing the product in the first term, we use the sum property ln(
x4y) – ln(7) = ln(x4) + ln(y) – ln(7)
Finally, we could use the exponent property on the first term ln(
x4) + ln(y) – ln(7) = 4ln(x) + ln(y) – ln(7)
David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, " Chapter 4: Exponential and Logarithmic Functions," licensed under a CC BY-SA 3.0 license.
Properties of Logs
Inverse Properties:
Exponential Property:
Change of Base:
While these properties allow us to solve a large number of problems, they are not sufficient to solve all problems involving exponential and logarithmic equations.
Properties of Logs
Sum of Logs Property:
Difference of Logs Property:
It's just as important to know what properties logarithms do
not satisfy as to memorize the valid properties listed above. In particular, the logarithm is not a linear function, which means that it does not distribute: log(A + B) ≠ log(A) + log(B).
To help in this process we offer a proof to help solidify our new rules and show how they follow from properties you've already seen.
Let
a = logb(A) and c = logb(C), so by definition of the logarithm, ba = A and bc = C
Using these expressions,
AC = babc
Using exponent rules on the right,
AC = ba+c
Taking the log of both sides, and utilizing the inverse property of logs, log
b(AC) logb(ba+c) = a + c
Replacing
a and c with their definition establishes the result logb(AC) = logbA + logbC
The proof for the difference property is very similar.
With these properties, we can rewrite expressions involving multiple logs as a single log, or break an expression involving a single log into expressions involving multiple logs.
Example 1
Write log 3(5) + log3(8) – log3(2) as a single logarithm. Using the sum of logs property on the first two terms, log 3(5) + log3(8) = log3(5 × 8) = log3(40) This reduces our original expression to log 3(40) – log3(2) Then using the difference of logs property,
Example 2
Evaluate 2log(5) + log(4) without a calculator by first rewriting as a single logarithm. On the first term, we can use the exponent property of logs to write 2log(5) = log(5 2) = log(25) With the expression reduced to a sum of two logs, log(25) + log(4), we can utilize the sum of logs property log(25) + log(4) = log(4 × 25) = log(100) Since 100 = 10 2, we can evaluate this log without a calculator: log(100) = log(102) = 2Try it Now 1
Without a calculator evaluate by first rewriting as a single logarithm: log 2(8) + log2(4)Example 3
Rewrite as a sum or difference of logs
First, noticing we have a quotient of two expressions, we can utilize the difference property of logs to write
Then seeing the product in the first term, we use the sum property ln(
x4y) – ln(7) = ln(x4) + ln(y) – ln(7)
Finally, we could use the exponent property on the first term ln(
x4) + ln(y) – ln(7) = 4ln(x) + ln(y) – ln(7)
Try it Now Answers
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David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, " Chapter 4: Exponential and Logarithmic Functions," licensed under a CC BY-SA 3.0 license.
