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
Solve for x: log6 x  2
Solution:
Let’s rewrite the problem
in exponential form.
62  x
We’re finished, x = 36
Example 2
1
Solve for y: log5
y
25
Solution:
Rewrite the problem
in exponential form.
1
5 
25
y
5y  5 2
y  2
 1
 2 
Since   5 
25

Example 3
Solve for x…
logx 54 = 4
Solution:
Try setting this up like this:
Rewrite in exponential form.
Now use your reciprocal power
to solve.
x4 = 54
(x4 )1/4= (54)1/4
x = (54)1/4
So x ≈ 2.7108
Example 4
Solve for x…
log3 (x+4) = 4
Solution:
Rewrite in exponential form.
Simplify the power expression.
34 = x + 4
81 = x + 4
Subtract 4 to finish…
x = 77
Example 5: More complicated expressions
Solve for x…
4 log (x+1) = 8
KEY KNOWLEDGE: You ALWAYS have to isolate the log expression
before you change forms.
Divide by the 4…
Rewrite in exponential form.
Remember common log is base 10.
Evaluate the power.
Subtract 1 to finish…
log (x+1) = 2
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: You ALWAYS have to isolate the log expression
before you change forms.
Add 6…
Multiply by 2 (or divide by 0.5)
0.5 log2 (3x+1) = 2
log2 (3x+1) = 4
Rewrite in exponential form.
.
Evaluate the power.
Subtract 1, then divide by 3 to
finish…
24 = 3x + 1
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.
Example 7
Solve:
log3 (4x 10)  log3 (x 1)
Solution:
Since the bases are both ‘3’ we simply set
the arguments equal.
Is the –3 a valid solution or is it
4x 10  x 1
3x 10  1
3x   9
x 3
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…
Example 8
Solve:
log8 (x 14)  log8 (5x)
2
Solution:
Since the bases are both ‘8’ we simply set the arguments equal.
2
x 14  5x
x2  5x 14  0
(x  7)(x  2)  0
Factor the quadratic
and solve
(x  7)  0 or (x  2)  0
continued on the next page
x  7 or x  2
Example 8 (cont)
Solve:
log8 (x 14)  log8 (5x)
2
Solution:
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. 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.
Example 9: Equations with more than one
log (Using our Condensing Skills)
Solve for x…
log2 (3x + 8) = log2 4 + log2 x
KEY KNOWLEDGE: You ALWAYS have to isolate the log expression
before you change forms. You may have to CONDENSE it!
Condense the right side …
log2 (3x+8) = log2 (4x)
Property of Equality…
Now solve... Subtract 3x
.
3x + 8 = 4x
8 = x
x=8
(Product Property)
Example 10: Equations with more than one
log (Using our Condensing Skills)
Solve for x…
log2 (3x) + log2 (8) = 4
KEY KNOWLEDGE: You ALWAYS have to isolate the log expression
before you change forms. You may have to CONDENSE it!
Condense the left side …
log2 (24x) = 4
Change into exponential form…
(Product Property… 3x*8 = 24x)
24 = 24x
Evaluate the power…
16 = 24x
Now solve... Divide by 24
and simplify
.
x = 16/24
x = 2/3
HOMEWORK:
WS – Solving Log Equations