4.3 Logarithmic Functions
In pairs: Individually, convert the following exponential equations to logarithmic form and their logarithmic
equations to exponential form. Then switch with your partner to check answers.
1. 5 = 7𝑡+2
3. 𝑦 = log 6 (7)
2. 8 = 7(3𝑥 ) − 4
4. 6 = 4(log 2 𝑥) − 9
In fours: Match Column I with Column II
I
II
Answer
1. 𝑙𝑜𝑔2 16
a. 0
1.
1
2.
2. 𝑙𝑜𝑔3 1
b.
3. 𝑙𝑜𝑔10 0.1
c. 4
3.
4. 𝑙𝑜𝑔2 √2
d. −3
4.
1
5. 𝑙𝑜𝑔𝑒 (𝑒 2 )
e. −1
5.
6. 𝑙𝑜𝑔1 8
f. −2
6.
2
2
In pairs: Individually, complete the missing
equivalent forms in the table. Then compare
answers with your partner. (we’ll call this #7)
34 = 81
a)
b)
log 1 8 = −3
2
103 = 1000
c)
d)
1
log 5 (125) = −3
121 = 12
e)
f)
log 6 (1) = 0
Individual:
8. Graph each function. Give the domain and range.
a) 𝑓(𝑥) = log 3 (𝑥 + 2)
b) 𝑓(𝑥) = log 2 (𝑥 − 3) + 2
In pairs: Compare answers from #8. Then graph the inverses of each function, give the equation of the inverse
function, the domain, and the range. Compare answers with your partner.
In fours: Use the properties of logarithms to expand (rewrite) the following expressions.
3
9. log 2 √9
5
10. log 𝑎 √𝑟 3
𝑚
𝑟 3 𝑠2
11. log 𝑎 √
𝑡4
In fours: Give the equation for each property, and a description of how that property behaves.
Property
Product Property
log 𝑎 (𝑥𝑦) = log 𝑎 (𝑥) + log 𝑎 (𝑦)
Quotient Property
𝑥
log 𝑎 ( ) = log 𝑎 (𝑥) − log 𝑎 (𝑦)
𝑦
Power Property
log 𝑎 (𝑥 𝑟 ) = 𝑟𝑙𝑜𝑔𝑎 (𝑥)
Logarithm of 1
log 𝑎 (1) = 0
Base a Logarithm of a
log 𝑎 𝑎 = 1
Conditions
for 𝑥 > 0, 𝑦 > 0, 𝑎 > 0, and 𝑎 ≠ 1
for 𝑥 > 0, 𝑦 > 0, 𝑎 > 0, and 𝑎 ≠ 1
for 𝑥 > 0, 𝑎 > 0, and 𝑎 ≠ 1
𝑎 > 0, and 𝑎 ≠ 1
𝑎 > 0, and 𝑎 ≠ 1
1. Solve each logarithmic equation.
a. 𝑙𝑜𝑔1 (𝑥 + 3) = 4
2
1 4
( ) =𝑥+3
2
1
=𝑥+3
24
1
=𝑥+3
16
1
𝑥=
−3
16
1 48
𝑥=
−
16 16
47
𝑥= −
≈ −2.9375
16
b. 𝑙𝑜𝑔𝑥+3 6 = 1
(𝑥 + 3)1 = 6
𝑥+3=6
𝑥=3
2. Write the following as one logarithmic expression by using the properties in the above table.
a. 𝑙𝑜𝑔𝑎 𝑥 + 𝑙𝑜𝑔𝑎 y − 𝑙𝑜𝑔𝑎 𝑚
log 𝑎 (𝑥𝑦) − log 𝑎 (𝑚)
𝑥𝑦
log a ( )
𝑚
b. 2𝑙𝑜𝑔𝑎 (𝑧 + 1) + 𝑙𝑜𝑔𝑎 (3𝑧 + 2)
log 𝑎 (𝑧 + 1)2 + log 𝑎 (3𝑧 + 2)
log 𝑎 [(𝑧 + 1)2 (3𝑧 + 2)]
log 𝑎 [(𝑧 + 1)(𝑧 + 1)(3𝑧 + 2)]
log 𝑎 [(𝑧 2 + 2𝑧 + 1) (3𝑧 + 2)]
log 𝑎 [3𝑧 3 + 8𝑧 2 + 7𝑧 + 2]
Why do you think that the base has to be the same for each of the expressions for us to be able to combine
the expressions?
The simple answer is because logs are exponentials and we need common bases in order to combine
exponentials.
Recall, 𝑦 = log 𝑎 (𝑥) can be rewritten as: 𝑎 𝑦 = 𝑥.
𝑎 𝑦 𝑎𝑟 = 𝑎 (𝑦+𝑟) , but 𝑎 𝑦 𝑏 𝑟 can’t be simplified further.
Whole Class: Talk about why these properties are useful.
Without these properties, some exponential equations can’t be solved. If the bases of exponential equations
can’t be rewritten in a way that gives us a common base, we would not be able to solve for 𝑥 without these
logarithmic properties.