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Pre-AP Precal
Teacher Notes
3.2 & 3.3 Notes Logarithmic Properties
2x = 20
How would you solve this problem? We know that it is a decimal. 24 = 16 and 25 =32, so it is
somewhere between 4 and 5. Just like we used inverse trig functions to solve trigonometric
equations, we need to use the inverse of an exponential, which is called a logarithm, to solve an
exponential equation.
We would solve this problem by converting it to logarithmic form:
log 2 20 = x (“log base 2 of 20 equals x”)
In general:
log b a = x
b is the base
a is the argument (or answer to an exponential equation)
x is the exponent
Common Logarithmic Function
base 10
log10x = log x
**If no base is indicated, it is base 10 (default base)
Logarithmic Form
Exponential Form
Logba = x
bx = a
Practice converting between log form and exponential form.
1. logb 5 = x
bx = 5
5
4. 2 = 32
log2 32 = 5
2. log3 81 = 4
3. log5 x = 3
34 = 81
53 =125
1
5. = 3-2
9
1
log3   = -2
9
1
4
6. 81 = 3
log81 3 =
1
4
Let’s use this knowledge to evaluate some logarithmic expressions.
7. log
1
100
First,convert to exponentialform.
1
100
Now convert to acommonbase.
10 x =
10 x =10-2
So, x=- 2
8. log4 64
4 x = 64
4 x = 43
x =3
9. log16
1
32
1
32
4x
2 = 2-5
16 x =
4x = -5
x=
-5
4
Properties of Logarithms
loga (x  y) = loga x +loga y
x
loga   = loga x - loga y
y
loga x y = yloga x
loga x =
logx
loga
Product Property
Quotient Property
Power Property
Change of base formula
logbb x = x
(I prove this one to them a couple of different ways.)
blogb x = x
(This one too.)
I also remind them here that logb 1 = 0.
Express the following as multiple logarithms.
10.
log a x 2 y 3 z 5  log a x 2  log a y 3  log a z 5
 2log a x  3log a y  5log a z
11.
xy 2
logb 3  logb x  logb y 2  logb z 3
z
 logb x  2logb y  3logb z
Express the following as a single logarithm. Simplify if possible.
12.
log519+log5 3 = log5 (19×3)
= log5 57
13.
3
1
3
1
loga x - loga y = loga x 2 - loga y 2
2
2
3
= loga
x2
y
= loga
14.
1
2
You could have them rationalize
this
x3
y
1
1
loga x + 3loga y - 2logaz = loga x 2 +loga y 3 - logaz 2
2
y3 x
= loga 2
z
Practice.
15. log51
16.
18. log2 4+log21- log2 8
19.
log
11
log7 5 7
11
17.
8log 8 30