# ALGEBRA II Lesson 10-3 ```ALGEBRA II
LESSON 10-3
Pg. 541-545
PROPERTIES OF LOGARITHMS
Today’s Key Concepts: We should know what a logarithmic function is, now we will work with the
multiplication, division, and exponential factors dealing with logarithmic functions.
Objectives:
(1) Simplify and evaluate expressions using the properties of logarithms
(2) Solve logarithmic equations using the properties of logarithms
Recall the EXPONENT lessons we had when multiplying, dividing, and power-to-a-power?
A. (x2) (x5) = x2+5 = x7 We added the exponents when multiplying.
B. (x14) = . (x14-3) = x11 We subtracted exponents when dividing.
(x3)
C. (x5)3 = x 5&middot;3 = x15 We multiplied exponents when power-to-a-power is shown
Logarithms are basically the same process!
The Three Laws of Logarithms
1. Product Property of Logarithms
logbxy = logbx + logby
&quot;The logarithm of a product (multiplication) is equal to the sum of the logarithms of each factor.&quot;
2. Quotient Property of Logarithms
logb x = logbx − logby
y
“The logarithm of a quotient (division) is equal to the logarithm of the numerator minus the logarithm
of the denominator.&quot;
3. Power Property of Logarithms
logb x n = n logbx
&quot;The logarithm of a power of x is equal to the exponent of that power (n) times the logarithm of x.&quot;
Demonstrate Examples of these Rules-----&gt;
EXAMPLE 1 (Law #1): (Product Property of Logarithms)
Given that the log5 2 ≈ 0.4307
NOTE: log5 2 = log10 2
Find log5 250
(Calculator Use) log10 5
Solution: log5 250 = log5 (125)(2) = log5 125 + log5 2
= log5 53 + log5 2
= 3 + 0.4307
= 3.4307
 125 = 53
≈ 0.4307
EXAMPLE 2 (Law #2) (Quotient Property of Logarithms)
Given log6 8 ≈ 1.1606 and log632 ≈ 1.9343
Find the value of log6 4
Solution: Since 32/8 = 4
Then log632
log6 8
 log632 - log6 8 ≈ 1.9343 - 1.1606
log6 4 ≈ 0.7737
EXAMPLE 3 (Law #3): (Power Property of Logarithms)
Given log5 6 ≈ 1.1133,
Approximate the value of log5 216
Solution: Since 216 = 63
log5 216 = log5 63
= 3 log5 6
≈ 3 (1.1133)
≈ 3.3399
________________________________________________________________________________
PROBLEMS: Solve each of these:
a) 4 log2 x – log2 5 = log2 125
x4 = 125  x4 = (125)(5) 
5
x4 = 625  x = 5
b) log8 x + log8 (x – 12) = 2 (Since “2” is not log8 2, we must follow the rule and state it as 82
x(x – 12) = 82 Product Property of Logarithms
x2 –12x = 64
x2 –12x – 64 = 0
(x –16)(x + 4) =0
x = 16
x ≠ - 4 (log8 –4 is undefined)
SAMPLE PROBLEMS:
#18 log5 30 = log5 (5)(2)(3) = log5 5 + log5 2 + log5 3 = 1 + 0.4307 + .6826 = 2.1133
#20 log5 10 = log5 10 - log5 9 = log5 (5)(2) - log5 (3)(3)
9
= (log5 5 + log5 2) – (log5 3 + log5 3)
= (1 + .4307) – (0.6826 + 0.6826)
= 0.0655
#26 2 log10 6 – 1/3 log10 27 = log10 x
62
271/3 = x
36 = x  x = 12
3
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