Logarithms
Logarithms are the inverse of exponentials. The exponential function 𝑦 = 𝑏 𝑥 , is a one-to-one
function1, so its inverse 𝑥 = 𝑏 𝑦 is a function. However written as 𝑥 = 𝑏 𝑦 , “x is a function of
y.” To express the inverse where “y is a function of x,” write 𝑦 = 𝑙𝑜𝑔𝑏 (𝑥). The statements
𝑥 = 𝑏 𝑦 and 𝑦 = 𝑙𝑜𝑔𝑏 (𝑥) are equivalent and both represent the inverse of 𝑦 = 𝑏 𝑥 .
Equivalent Exponential and Logarithmic Forms
𝑥 = 𝑏 𝑦 is equivalent to 𝑦 = 𝑙𝑜𝑔𝑏 (𝑥)
* For any positive base b, where b≠1.
The above exponential statement, 𝑥 = 𝑏 𝑦 is equivalent to the logarithmic statement, 𝑦 =
𝑙𝑜𝑔𝑏 (𝑥), and is pronounced, “y equals log base b of x.” The value of the subscript “b” is the
“base of the logarithm,” just as “b” is the base of the exponential in the expression “𝑏 𝑦 .”
The below picture shows 𝑦 = 10𝑥 , and its inverse 𝑦 = log10 𝑥. As shown on the graph, the
base number “10” is omitted and the logarithm is written as 𝑦 = log 𝑥. Logarithms with base
10 are considered common logs. Common logs can be written without showing the 10.
Warning: economists usually mean natural log (a log with base 𝑒) and NOT the base-ten log
when they write the log without a base.
Original Exponential Function
(Eq1)
𝑦 = 10𝑥
Inverse of Original Function
In Exponential Form:
(Eq2)
𝑥 = 10𝑦
(Eq3)
In Logarithmic Form:
𝑦 = log10 𝑥
The remainder of this less will focus on writing equivalent exponential and logarithmic
statements, rather than working between an original statement and its inverse. In effect,
you will practice the skills necessary to transform the above Eq2 into Eq3.
1
𝑓
𝑓 −1
Definitions
Let 𝑏 > 0,
D1. If 𝑥 > 0, then the logarithm of 𝑥 to the base 𝑏 denoted log 𝑏 𝑥 , is the number 𝑦 such
that 𝑏 𝑦 = 𝑥; that is,
𝐲 = 𝐥𝐨𝐠 𝐛 𝐱 𝐢𝐟 𝐚𝐧𝐝 𝐨𝐧𝐥𝐲 𝐢𝐟 𝒃𝒚 = 𝒙 𝐟𝐨𝐫 𝒙 > 𝟎.
In other words, log 𝑏 𝑥 is the exponent you put on 𝑏 to give you 𝑥:
log 2 (8) ⇒ 2𝑦 = 8 = 23 ⇒ 𝑦 = 3.
⇒ log 2(8) = 3
In this case, log 2 (8) yielded the corresponding power to which 2 would have to be
raised in order to get 8, which is 3 because 23 = 8.
D2. log 𝑏 (𝑏 𝑥 ) = 𝑥 for all 𝑥 and 𝑏 log𝑏 𝑥 = 𝑥 for all 𝑥 > 0.
log 3 27 = log 3 (33 ) ⇒ 3𝑥 = 33 ⇒ 𝑥 = 3.
⇒ log 3 (33 ) = 3.
D3. ln 𝑥 = log 𝑒 𝑥, where 𝑒 ≈ 2.718. In this case, the natural logarithm of 𝑥 is the power
to which 𝑒 would have to be raised in order to get 𝑥.
ln 20 = log 𝑒 20 ⇒ 𝑒 𝑥 = 20 ⇒ 𝑒 3 ≈ 20 ⟹ 𝑥 ≈ 3 .
⇒ ln 20 ≈ 3
In this case, ln 20 yielded the corresponding power to which 𝑒 would have to be
raised in order to get 20, which is approximately 3 because 𝑒 3 ≈ 20.
D4. ln 𝑒 = log 𝑒 𝑒 = 1 and ln 1 = log 𝑒 1 = 0.
D5. 𝑒 ln 𝑥 = 𝑥 for 𝑥 > 0 and ln 𝑒 𝑥 = 𝑥 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥.
Theorem (Laws of Logarithms)
Let 𝑏 > 0. For any 𝑥, 𝑦 > 0, we have:
LL0.
log 𝑏 𝑥 = log 𝑏 𝑦 if and only if 𝑥 = 𝑦 (𝐿𝐿0 Equality)
LL1.
log 𝑏 (𝑥 ∙ 𝑦) = log 𝑏 𝑥 + log 𝑏 𝑦
LL2.
log 𝑏 (𝑦) = log 𝑏 𝑥 − log 𝑏 𝑦
LL3.
log 𝑏 (𝑥 𝑎 ) = 𝑎 ∙ log 𝑏 𝑥
LL4.
log 𝑏 (𝑏 𝑥 ) = 𝑥
𝑥
(𝐿𝐿1 Product)
(𝐿𝐿2 Quotient)
(𝐿𝐿3 Power)
(LL4 Inversion)
LL5.
log 𝑏 𝑥 =
log𝑎 𝑥
log𝑎 𝑏
(𝐿𝐿5 Change of base formula)
*Note that log 𝑏 (𝑥 + 𝑦) ≠ log 𝑏 𝑥 + log 𝑏 𝑦 and
log𝑏 𝑥
log𝑏 𝑦
≠ log 𝑏
𝑥
𝑦
𝑥
𝑦
and log 𝑏 (𝑥 − 𝑦) ≠ log 𝑏 ( )
log 2 16 = 4 ≠ log 2 8 + log 2 8 = 3 + 3 = 6
Concepts for Graphing (not included in practice problems)
Logarithmic Concept Summary
Parent Function
𝑦 = log 𝑏 𝑥, 𝑏 > 0, 𝑏 ≠ 1
Stretch (|𝑎| > 1)
Compress (shrink) (0 < |𝑎| < 1)
𝑦 = alog 𝑏 𝑥
Reflection (𝑎 < 0) in x-axis
Translations (horizontal by h; vertical by k)
All Transformations combined
𝑦 = log 𝑏 (𝑥 − ℎ) + 𝑘
𝑦 = alog 𝑏 (𝑥 − ℎ) + 𝑘
* With a function 𝑦 = 𝑎log 𝑏 𝑥, where a>0 the domain is (0, ∞) and the range is (−∞, ∞).
Note the impact of the following transformations when made in isolation.




If a<0, the function will flip over the x-axis but will not change the domain or range.
If “x” is negated in 𝑦 = 𝑎log 𝑏 𝑥, i.e. 𝑦 = 𝑎log 𝑏 (−𝑥), the function will flip over the y-axis
and make the domain (−∞, 0).
Adding “k” will vertically shift the function, but will not change the domain or range.
Adding “h” will horizontally shift the function and make the domain (ℎ, ∞).
Examples
1. log 3 81
Remember that log 3 81 is the exponent we put on 3 to get 81. Using D1 we have that
log 3 81 implies that 3𝑥 = 81 , (with 𝑥 = log 3 81)
⇒ Since we know that 34 = 81, then x = log 3 81 = 4.
2. log12 144
Using D1 we have that log12 144 implies that 12𝑥 = 144 ⇒ since we know that 122 =
144, then log12 144 = 2.
1
3. log 3 27
Using the law of exponents, we know that
log 3
1
27
1
27
=
1
33
= 3−3. Using D1 and D2 we have that
= log 3 3−3 = −3.
1
4. log10 (1000)
1
Remember that log10 (1000) is the exponent we put on 10 to get 1/1000. Using D1 we
1
1
have that log10 (1000) ⇒ 10𝑥 = 1000 ⇒ since we know that 10−3 = 1/1000, then
1
log10 (1000) = −3.
5. log 5 1252
Using the law of exponents, we know that 1252 = (5 ∙ 5 ∙ 5)2 = (53 )2 = 56 . Using D1
and D2 we have that log 5 1252 = log 5 56 = 6.
6. 3log3 9
Using D2 we know that 3log3 9 = 9. Since log 3 9 = 2, then 32 = 9.
7. 5− log5 (25)
1
Using the law of exponents we have that, 5− log5 (25) = 5log5(25) . Using D1 and D2 we have
1
1
1
that 5− log5 (25) = 5log5(25) = 52 = 25.
8. log 4 6410
Using the law of exponents, we know that 6410 = (4 ∙ 4 ∙ 4)10 = (43 )10 = 430 .
Using D1 and D2 we have that log 4 6410 = log 4 430 = 30.
9. log 2 410
Using the law of exponents, we know that 410 = (2 ∙ 2)10 = (22 )10 = 220 .
Using D1 and D2 we have that log 2 410 = log 2 220 = 20.
5
10. log 2 √8
5
Using the law of exponents, we know that √8 = 81/5 = (2 ∙ 2 ∙ 2)1/5 = (23 )1/5 = 23/5 .
3
5
Using D1 and D2 we have that log 2 √8 = log 2 (25 ) = 3/5.
11. log √6 36
2 2
4
Using the law of exponents, we know that 62 = 36, then ((√6) ) = √6 . Hence,
4
log √6 36 = log √6 √6 = 4.
12. log10
1
√10
Using the law of exponents we know that,
have that log10
1
√10
1
1
√10
=
1
101/2
= 10−1/2 . Using D1 and D2 we
= log10 (10−2 ) = −1/2.
13. log10 1,000,000 + log10 1,000 − log10 0.1
Using D1 we have that log10 1,000,000 implies that 10𝑥 = 1,000,000 ⇒ 𝑥 = 6. We also
know that log10 1,000 implies that 10𝑥 = 1,000 ⇒ 𝑥 = 3. Finally, log10 0.1 implies that
10𝑥 = 0.1 ⇒ 𝑥 = −1 because
1
10
= 0.1 = 10−1 .
Hence, log10 1,000,000 + log10 1,000 − log10 0.1 = 6 + 3 − (−1) = 10.
14. log 2
16𝑥 3
7
Using LL2 and LL3 we have that, log 2
16𝑥 3
7
= log 2 16𝑥 3 − log 2 7 = log 2 16 +
log 2 𝑥 3 − log 2 7 = 4 + 3 log 2 𝑥 − log 2 7.
27(𝑥+1)3
15. log 3 𝑥 2 (𝑥−3)4
27(𝑥+1)3
Using LL1, LL2 and LL3 we have that, log 3 𝑥 2 (𝑥−3)4 = log 3 27(𝑥 + 1)3 − log 3 𝑥 2 (𝑥 −
3)4 = log 3 27 + 3 log 3 (𝑥 + 1) − 2 log 3 𝑥 − 4 log 3 𝑥 − 3 = 3 + 3 log 3 (𝑥 + 1) − 2 log 3 𝑥 −
4 log 3 𝑥 − 3.