CHAPTER 2
LESSON 8
Teacher’s Guide
Equations involving logarithms
AW 2.3, 2.12
MP 2.7
Objective:
• To solve equations involving logarithms
Example 1:
Solve for x: log 2 (x − 2) + log2 x = log 2 3
Principle:
If log b p = log b q
then p = q.
G-210
Using the law of logarithms for multiplication we have
log 2 (x − 2) ⋅ x = log 2 3.
Since both sides are logs to the same base, the equality can be true only if
(x − 2)⋅ x = 3
2
x − 2x − 3 = 0
(x − 3)(x + 1) = 0
x = 3 or x = −1.
Check the answers in the original equation.
x=3
LHS
log 2 (x − 2) + log2 x
= log2 (3 − 2) + log 2 3
= log2 1 + log 2 3
= log2 3
RHS
log 2 x
= log 2 3
x = −1
Since log 2 ( −1) on the right hand side does not exist, x = −1 must be rejected as an
answer.
Therefore the only solution is x = 3.
Once you have solved a logarithmic equation, you must check each value of your
solution. Substitute each value into the original equation, and make sure that the
equation is defined for this value of the variable. In other words, if you find a value for
which the original statement of the equation is undefined – forcing you to take the log of
a negative number, or zero – then you must reject that value as part of the solution.
Example 2:
log (x − 1) + log(2x − 3) = log(2x 2 − 5)
Solve for x:
Using the law of logarithms for multiplication,
log (x − 1)⋅ (2x − 3) = log (2x 2 − 5)
Since both sides are logs to the same base, the equality can be true only if:
(x − 1) ⋅ (2x − 3) = 2x 2 − 5
2x 2 − 5x + 3 = 2x 2 − 5
−5x = −8
x = 8 5=1.6
Check:
LHS
RHS
log(x − 1) + log(2x − 3)
log(2x 2 − 5)
= log(1.6 − 1) + log[2 ⋅ (1.6) − 3]
= log0.6 + log 0.2
= log0.12
= log[2 ⋅ (1.6)2 − 5 ]
= log 0.12
Example 3:
Solve for x, checking for any extraneous solutions.
G-210
log 5 (3 x + 1) + log 5 ( x − 3) = 3
log 5 (3x + 1)(x − 3) = 3
log 5 (3x 2 − 8x − 3) = 3
2
3
3x − 8x − 3 = 5
2
3x − 8x − 3 = 125
(change from log statement to exponential
statement)
2
3x − 8x − 128 = 0
(3x + 16)(x − 8) = 0
x = − 16 3 or x = 8
Check:
x = − 16 3
LHS
RHS
log 5 (3x + 1) + log 5 ( x − 3)
3
= log5 [3(−16 3) + 1]+ log 5 [−16 3 + 1]
We reject x = − 16 3 because it means taking the log of a negative number.
x=8
LHS
log 5 (3x + 1) + log 5 ( x − 3)
= log5 (3 ⋅ 8 + 1) + log 5 (8 − 3)
= log5 (25 ) + log 5 (5)
= 2+ 1
=3
The only solution is x = 8.
RHS
3
Example 4:
Given log a 2 = x and (log a 8)(a loga x ) = 12 , solve for a.
(Exam Specs #7 p. 16)
First, we have loga 8 = log a 23 = 3log a 2 = 3x.
Also, a log a x = x
So, 3x ⋅ x = 12
x2 = 4
x = 2 (Reject x = −2)
Finally,
x = loga 2 = 2
2
a =2
a = 2 (Reject a = − 2 )
Example 5:
Solve for x: log a a 2x = log b 2 b3x − 3
Hint: recall law 2 of lesson 6:
(#21, June 2001 Provincial Exam)
log b n x n = logb x
n ∈ℜ, b > 0, b ≠ 1, x > 0
Using the extension of the change of base law, we have
1
2x = log b[b 3x − 3 ]2
2x = log bb
1 (3x− 3)
2
2x = 1 (3x − 3)
2
x = −3
G-210
Example 6:

Verify the identity log a  1
x  = − log a x for any base a and any positive value of x.
LHS

log a  1
x
= log a 1− log a x
= 0 − loga x
= − log a x
RHS
− log a x
Example 7:
Solve for x:
log 4 2x + 2 − log 4 3x + 1 = 1
2
First, use the law of logarithms for division.
2x + 2 = 1
log 4 3x
+1 2
+ 2 = 4 12 = 2
Therefore, 2x
3x + 1
2x + 2
Then, either 3x
+ 1 = 2 or
2x + 2
3x + 1 = −2
2x + 2 = 2
3x + 1
2x + 2 = 6x + 2
2x + 2 = −2
3x + 1
2x + 2 = −6x − 2
x=0
x=− 1
2
Example 8:
Graph log 5 (y + 2) = x + 1 on the grid below. State any asymptotes and give exact values
for the x- and y- intercepts.
First, change the log equation to an exponential equation.
y + 2 = 5 x +1
y = 5 x+ 1 − 2
How does the graph of this equation compare to the graph of y = 5 x ?
Horizontal Shift: 1 unit to the left________________
Vertical Shift: 2 units down_____________________
7
6
5
4
3
2
1
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
-1
-2
-3
-4
Asymptote: ______y = –2______________
x-intercept (zero):
0 = 5 x +1 − 2
5
x+ 1
Set y = 0 .
=2
(x + 1)log5 = log2
x +1=
x=
log2
log5
log2
− 1= log5 2 − 1 ( =⋅ − 0.5693)
log5
y-intercept: Set x = 0 .
y = 5 0+1 − 2 = 3