Download Lesson 26

Algebra III
Lesson 26
The Logarithmic Form of the Exponential – Logarithmic Equations
The Logarithmic Form of the Exponential
N = bL
N = answer; b = the base; L = power or exponent
The exponent of b is L.
The power of b is L.
The logarithm for b to get N is L.
The logarithm function is used to pull down the exponent, not
to undo the power.
N = bL
logb N = L
The log base b of N is L
Example 26.1
Write 4 = 3y in the logarithmic form.
Recall:
N = bL
So 4 = 3y becomes
logb N = L
log3 4 = y
Example 26.2
Write logx 4 = m in exponential form.
N = bL
logb N = L
So logx 4 = m becomes
4 = xm
Logarithmic equations
Two key things to watch for when solving these:
1st – Generally make sure the problem is written in
exponential form not logarithmic form.
2nd – Many times working both sides of the exponential
equation in terms of the same base raised to powers.
Example: 16 = 2r
16 = 24
24 = 2r
Therefore, r = 4.
Example 26.3
Solve: logb 8 = 3
1st rewrite in exponential form
N = bL
logb N = L
So logb 8 = 3 becomes:
8 = b3
This is now a problem we know how to deal with.
Take the cube root of both sides.
(8 = b3)⅓
8⅓ = b
2=b
Example 26.4
Solve: logb 9 = - ½
Rewrite:
9 = b-½
Square both sides to clear the root.
(9 = b-½)2
81 = b-1
Take both sides to the -1 power to clean up the b.
(81 = b-1)-1
1/81 = b
Example 26.5
Solve:
log 3
1
=M
27
Rewrite:
1
= 3M
27
In the previous problems once it was rewritten it liked similar to
things we had already done. All that was needed was to take a
root. This is different.
In this case we need to get both sides in terms of the same
number to a power.
Since one side is 3M lets try to get the other side as a power of 3.
Start with the 27.
27 = 33
So, 1/27 = 1/33
1/33is not quite in terms of 3 to a power.
1/33 = 3-3
This is 3 to a power.
So, 1/27 = 1/33 = 3-3
Therefore, if 3-3 = 3M, then M = -3
Example 26.6
Solve:
log 1 P = −2
3
Rewrite:
−2
1
  =P
 3
This isn’t to bad to do.
−2
1
P =   = 32 = 9
 3
Example 26.7
Solve: log4 8 = X
Rewrite:
8 = 4X
Both sides need to becomes some number taken to a
power, and it needs to be the same number for both sides.
Since, 8 cannot be found by taking 4 to a nice power,
what can both sides be converted to as bases?
Try powers of 2.
8 = 23
4X = (22)X = 22X
So, 8 = 4X becomes 23 = 22X.
Which makes 2X = 3, therefore X = 2/3.
Practice
a) 1] Write kP = 7 in logarithmic form.
N = bL
logb N = L
logk 7 = P
2] Write logk 7 = P in exponential form.
kP = 7
b) Solve: logb 27 = 3
Rewrite:
b3 = 27
Now cube root both sides.
(b3 = 27)⅓
b=3
c) Solve: log 2
1
=M
8
Rewrite:
1
M
 =2
8
Since the variable is the power, both sides need to be in
terms of the same base. And that is going to be 2.
Examine the 8, can this be a power of 2?
8 = 23
So,
1 1
 = 3
8 2
which becomes 2-3.
Therefore, 2M = 2-3, making
M = -3
d) Solve: log 1 C = −4
2
Rewrite:
−4
1
  =C
 2
24 = C
C = 16
e) Factor: 64x12y6 – 27a6b9
Squares?
Are the numbers perfect squares?
Are all the powers even?
No.
Cubes?
Are the numbers perfect cubes?
Are all the powers multiples of 3?
(64 x
12
) ((
) (
3
y 6 − 27 a 6b 9 = 4 x 4 y 2 − 3a 2b 3
Use the correct pattern:
(
)(
))
Yes.
3
a3 - b3 = (a - b) (a2 + ab + b2)
= 4 x 4 y 2 − 3a 2b 3 16 x 8 y 4 + 12a 2b 3 x 4 y 2 + 9a 4b 6
)