1. Laws of Logarithms
Since logarithms are exponents, the Laws of Exponents give rise to the Laws of Logarithms.

2. Example 1 – Using the Laws of Logarithms to Evaluate Expressions
Evaluate each expression.
(a) log4 2 + log4 32
(b) log2 80 – log2 5
(c) log 8
Solution: (a) log4 2 + log4 32 = log4(2  32)
= log4 64 = 3
Law 1
Because 64 = 43

3. Example 1 – Solution (b) log2 80 – log2 5 = log2 = log216 = 4
cont’d
(b) log2 80 – log2 5 = log2
= log216 = 4
(c) log 8 = log 8–1/3
= log
 –0.301
Law 2
Because 16 = 24
Law 3
Property of negative exponents
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4. Expanding and Combining Logarithmic Expressions
The Laws of Logarithms allow us to write the logarithm of a product or a quotient as the sum or difference of logarithms.
This process, called expanding a logarithmic expression

5. Example 2 – Expanding Logarithmic Expressions
Use the Laws of Logarithms to expand each expression.
(a) log2(6x) (b) log5(x3y6) (c) ln
Solution: (a) log2(6x) = log2 6 + log2 x
(b) log5(x3y6) = log5 x3 + log5 y6
= 3 log5 x + 6 log5 y
Law 1
Law 1
Law 3

6. Example 2 – Solution (c) = ln(ab) – = ln a + ln b – ln c1/3
cont’d
(c) = ln(ab) –
= ln a + ln b – ln c1/3
= ln a + ln b – ln c
Law 2
Law 1
Law 3

7. Expanding and Combining Logarithmic Expressions
The Laws of Logarithms also allow us to reverse the process of expanding
That is, we can write sums and differences of logarithms as a single logarithm.
This process, called combining logarithmic expressions, is illustrated in the next example.

8. Example 3 – Combining Logarithmic Expressions
Combine 3 log x + log (x + 1) into a single logarithm.
Solution: 3log x + log(x + 1) = log x3 + log(x + 1)1/2
= log(x3(x + 1)1/2)
Law 3
Law 1

9. Expanding and Combining Logarithmic Expressions
Logarithmic functions are used to model a variety of situations involving human behavior.
One such behavior is how quickly we forget things we have learned.
For example, if you learn algebra at a certain performance level (say, 90% on a test) and then don’t use algebra for a while, how much will you retain after a week, a month, or a year?

10. Example 5 – The Law of Forgetting
If a task is learned at a performance level P0, then after a time interval t the performance level P satisfies
log P = log P0 – c log(t + 1)
where c is a constant that depends on the type of task and t is measured in months.
(a) Solve for P.
(b) If your score on a history test is 90, what score would you expect to get on a similar test after two months? After a year? (Assume that c = 0.2.)

11. Example 5 – Solution (a) We first combine the right-hand side.
log P = log P0 – c log(t + 1)
log P = log P0 – log(t + 1)c
log P =
P =
Given equation
Law 3
Law 2
Because log is one-to-one

12. Example 5 – Solution
cont’d
(b) Here P0 = 90, c = 0.2, and t is measured in months.
In two months: t = and
In one year: t = and
Your expected scores after two months and one year are and 54, respectively.

13. Change of Base Formula
to change from logarithms in one base to logarithms in another base. y = logb x We write this in exponential form and take the logarithm, with base a, of each side.
by = x Exponential form
loga (by) = loga x Take loga of each side
yloga b = loga x Law 3
Divide by logab

14. Change of Base Formula This proves the following formula.
In particular, if we put x = a, then loga a = 1, and this formula becomes

15. Example 6 – Evaluating Logarithms with the Change of Base Formula
Use the Change of Base Formula and common or natural logarithms to evaluate each logarithm, correct to five decimal places.
(a) log (b) log920
Solution:
(a) We use the Change of Base Formula with b = 8 and a = 10:
log8 5 = 

16. Example 6 – Solution
cont’d
(b) We use the Change of Base Formula with b = 9 and a = e:
log9 20 = 