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Logarithms
A logarithm of a given number 𝑥, is the exponent required for the base 𝑎, to be raised to in
order to produce that number 𝑥.
log
𝑎
𝑦
Note that ⇔ means "𝑖𝑠 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑡𝑜"
Logarithmic and Exponential Form
Change logarithm equations to exponential form or exponential equations to logarithmic
form using the definition of a logarithm.
Example: Given 4
3
2
⁄
= 8 , change the equation to logarithmic form.
Solution:
Compare the equation to the definition and rewrite it.
Definition: log 𝑎
𝑦
Given: 4
3
2
⁄
Notice that 𝑎 = 4 , 𝑥 = 8 , and
, respectively.
Therefore, using the definition: 4
3
2
⁄
𝟒
Example: Given log
25
, change the equation to exponential form.
Solution:
Compare the equation to the definition and rewrite it.
Definition: log
𝑎
𝑦
Given: log 25
Notice that 𝑎 = 25 , 𝑥 = 5 , and
, respectively.
Therefore, using the definition: log
25
𝟏
𝟐
⁄
Solving Logarithm and Exponential Equations
Evaluate logarithmic equations by using the definition of a logarithm to change the
equation into a form that can then be solved.
Example: Given 3
𝑥− 1
= 7 , solve for 𝑥.
Solution:
Step 1: Set up the equation and use the
definition to change it.
Definition: log 𝑎
𝑦
Given 3
𝑥− 1
Notice 3 is the base or 𝑎,
and 7 is the given number.
𝑥− 1
= 7 ⇔ log
3
Step 2: Now use the properties of
logarithms to solve.
Recall the Change of Base Property:
log
𝑎
log 𝑏
log 𝑎
Apply it to log
3
log
3
log 7
log 3
Step 3: Use the order of operations to finish solving for 𝑥.
log 7
log 3
Example: Given log
6
(𝑥 + 2 ) = 3 , solve for 𝑥.
Solution:
Step 1: Set up the equation and use the
definition to change it.
Definition: log
𝑎
𝑦
Given log
6
Notice 6 is the base or 𝑎,
and 3 is the exponent or 𝑦.
log 6
3
Step 2: Now use the order of
operations to solve.
3
Solving Expanded Logarithms
Solving expanded logarithms requires applying the definition of logarithms and all the
logarithm properties as needed.
Example: Given ln(𝑥 − 2 ) + ln(𝑥 − 3 ) = ln( 2 𝑥 + 24 ) , solve for 𝑥.
Solution:
Note: ln(𝑥 − 2 ) is only valid if 𝑥 ≥ 2 , ln(𝑥 − 3 ) is only valid if 𝑥 ≥ 3 , and ln( 2 𝑥 + 24 )
is only valid if 𝑥 ≥ − 12. For the equation to be valid, all conditions must be met, so 𝑥 ≥ 3.
Step 1: Simplify the left side of the
equation using the multiplication and
division properties of logarithms.
ln(𝑥 − 2 ) + ln(𝑥 − 3 ) = ln( 2 𝑥 + 24 )
⇒ ln(𝑥 − 2 )(𝑥 − 3 ) = ln( 2 𝑥 + 24 )
⇒ ln(𝑥
2
− 5 𝑥 + 6 ) = ln( 2 𝑥 + 24 )
Step 2: Use logarithm properties.
Recall logarithm properties of bases:
ln 𝑒
𝑥
= 𝑥 and 𝑒
ln 𝑥
ln(𝑥
2
− 5 𝑥 + 6 ) = ln( 2 𝑥 + 24 )
Let both sides of the equation become the
exponent of the base 𝑒, and apply the
property.
ln(𝑥
2
− 5 𝑥+ 6 )
ln( 2 𝑥+ 24 )
2
Step 3: Combine like terms to solve for 𝑥.
2
2
Step 4: Check your answers. Recall that
every logarithm must meet the conditions
for the answer to be correct.
For 𝑥 = 9
ln(( 9 ) − 2 ) + ln(( 9 ) − 3 ) = ln( 2 ( 9 ) + 24 )
⇒ ln( 7 ) + ln( 6 ) = ln( 42 )
⇒ ln( 7 ⋅ 6 ) = ln( 42 ) ⇢ This is valid!
For 𝑥 = − 2
Since − 2 ≱ 3 , it does not meet all the
conditions, and is not valid.
Therefore: 𝒙 = 𝟗
Practice Exercises:
- Given log 4
4
Solve for x.
- Expand log
2
2
) completely.
- Write the following as a single
logarithm: 2 log 3
𝑥 + 4 − 8 log
3
Answers:
- log
2
log
2
log
2
- log
3
3
8
