Solving Logarithmic Equations
(Day 1)
Wednesday, March 1, 2017
Warm-Up
• Factor:
𝑥𝑥 2 − 3𝑥𝑥 − 40
Essential Question
• How can I use exponential form to help me solve logarithmic
equations?
Solving equations with logarithms:
We now know that a logarithm is perhaps best understood as being
closely related to an exponential equation.
In fact, whenever we get stuck in the problems that follow we will
return to this one simple insight.
We might even state it as a simple rule.
When working with logarithms, if ever you get “stuck”, try rewriting
the problem in exponential form.
Conversely, when working with exponential expressions, if ever you
get “stuck”, try rewriting the problem in logarithmic form.
Example 1
Solution:
Solve for x: log 6 x = 2
Let’s rewrite the problem
in exponential form.
6 = x
2
We’re finished, x = 36
Example 2
Solution:
y = −2
1
Solve for y: log 5
=y
25
Rewrite the problem
in exponential form.
1
5 =
25
y
5 =5
y
Alternate: If you can
evaluate the log, you will
already have the value of
the y!
log5 (1/52) = log5 5-2 = -2
 1
− 2
Since  = 5 


25
−2
Example 3
Solve for x…
Try setting this up like this:
Solution:
Rewrite in exponential form.
Now use the reciprocal
power to solve.
logx 54 = 4
x4 = 54
(x4 )1/4= (54)1/4
x = (54)1/4
So x ≈ 2.7108
Solve for x…
Example 4
log3 (x+4) = 4
Rewrite in exponential form.
34 = x + 4
Simplify the power expression.
81 = x + 4
Solution:
Subtract 4 to finish…
x = 77
Example 5: More complicated expressions
KEY KNOWLEDGE:
You ALWAYS have to
isolate the log
expression before
you change forms.
Solve for x…
4 log (x+1) = 8
Divide by the 4…
log (x+1) = 2
Rewrite in exponential form.
Remember common log is base 10.
Evaluate the power.
Subtract 1 to finish…
102 = x + 1
100 = x + 1
x = 99
Example 6: More complicated expressions
Solve for x…
0.5 log2 (3x+1) – 6 = –4
KEY KNOWLEDGE:
Add 6…
0.5 log2 (3x+1) = 2
You ALWAYS have
to isolate the log Multiply by 2 (or divide by 0.5)
log2 (3x+1) = 4
expression before
4 = 3x + 1
Rewrite in exponential form.
2
you change forms. .
Evaluate the power.
Subtract 1, then divide by 3 to finish…
16 = 3x + 1
15 = 3x
x=5
Finally, we want to take a look at the Property of Equality for Logarithmic Functions.
Suppose b > 0 and b ≠ 1.
Then logb x1 = log b x 2 if and only if x1 = x 2
Basically, with logarithmic functions, if the bases match on both
sides of the equal sign , then simply set the arguments equal to
each other.
Example 7
Solution:
Solve:
log3 (4x +10) = log3 (x +1)
Since the bases are both ‘3’ we
simply set the arguments equal
to each other and solve.
Is the –3 a valid solution or is it extraneous??
Note: If I plug -3 into the original, it causes me to take
the log of a negative number!!!
Therefore, it is NOT valid. Our answer is NO
SOLUTION…
4x +10 = x +1
3x +10 = 1
3x = − 9
x= −3
Example 8
Solution:
Solve:
log8 (x −14) = log8 (5x)
2
Since the bases are both ‘8’ we
simply set the arguments equal.
x −14 = 5x
2
x − 5x −14 = 0
2
Factor the quadratic
and solve
(x − 7)(x + 2) = 0
(x − 7) = 0 or (x + 2) = 0
x = 7 or x = −2
continued on the next page…
Example 8 (cont)
Solution:
Solve:
log8 (x −14) = log8 (5x)
2
x = 7 or x = −2
It appears that we have 2 solutions here.
If we take a closer look at the definition of a logarithm however, we will see
that not only must we use positive bases, but also we see that the
arguments must be positive as well.
Therefore, -2 is not a solution because it causes the log of a negative #.
Each side becomes the log8 (-10) !!!!!
So that answer is extraneous and has to be thrown out.
Only x = 7 works in the original equation… log 8 35 on both sides!
Solve for x…
log2 (3x + 8) = log2 4 + log2 x
Condense the right side using Product Property…
Example 9: Equations with more
than one log (Using our Condensing
Skills)
log2 (3x+8) = log2 (4x)
Property of Equality…
3x + 8 = 4x
Now solve... Subtract 3x
.
8 = x
x=8
KEY KNOWLEDGE:
You ALWAYS have
to isolate the log
expression before
you change forms.
You may have to
CONDENSE it!
Solve for x…
log2 (3x) + log2 (8) = 4
Condense the left side …
log2 (24x) = 4
Change into
exponential form…
24
= 24x
16 = 24x
Now solve... Divide by 24 x = 16/24
and simplify the fraction.
x = 2/3
Evaluate the power…
Example 10: Equations
with more than one log
(Using our Condensing
Skills)
(Product Property… 3x*8 = 24x)
KEY KNOWLEDGE:
You ALWAYS have to
isolate the log
expression before you
change forms. You
may have to
CONDENSE it!
HOMEWORK:
WS: Solving Log
Equations