Mathematics Skill Development - Module 5
Mathematics Skill Development - Module 5
Exponential and Logarithmic Functions
The following questions will evaluate the student’s understanding of logarithm properties. Moreover,
there will be emphasis on the relationship between logarithmic and exponential functions.
1. Evaluate the following logarithms:
(a) log2 64.
(b) log10 1000.
(c) ln e100 .
Solution:
(a) Let y = log2 64. Since log2 (x) is the inverse function of 2x , we have 2y = 64 and y = 6.
Alternatively, 64 = 26 and log2 64 = log2 26 = 6 log2 (2) = 6 by the properties of logarithms.
(b) Let y = log10 1000. Since log1 0(x) is the inverse function of 10x , we have 10y = 1000 and y = 3.
Alternatively, we write 1000 = 103 and log10 (1000) = log10 103 = 3 log10 (10) = 3 by the properties of
logarithms.
(c) Since ln(x) is the inverse function of ex , ln (ex ) = x for every x. In particular ln e
100
= 100.
2. Simplify the expression
e
1
2
2 ln ln ex
.
Solution:
2
Sine ln (ex ) = x for every number x, ln ex = x2 . Using the properties of logarithms,
1
1
ln x2 = · 2 ln x = ln x.
2
2
Therefore,
1
e2
2 ln ln ex
= eln x = x.
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Mathematics Skill Development - Module 5
3. Solve the following equations:
(a) 2ex = 6x+3 + ln(1).
(b) −2 ln x + ln x4 = ln(2x − 1).
Solution:
(a) Since ln(1) = 0, we get
2ex = 6x+3 + ln(1)
2ex = 6x+3 .
Taking the natural logarithm of both sides and using the properties of logarithms gives
ln(2ex ) = ln 6x+3
ln 2 + ln(ex ) = (x + 3) ln 6
ln 2 + x = (ln 6)x + 3 ln 6
x(1 − ln 6) = 3 ln 6 − ln 2
3
6
x(1 − ln 6) = ln
2
3
6
ln 2
x=
1 − ln 6
ln 108
=
.
1 − ln 6
(b) Using properties of the logarithm, we rearrange the equation as
−2 ln x + ln x4 = ln(2x − 1)
ln x−2 + ln x4 = ln(2x − 1)
ln x−2 · x4 = ln(2x − 1)
ln(x2 ) = ln(2x − 1).
Exponentiating both sides, we have
x2 = 2x − 1
x2 − 2x + 1 = 0.
This is a quadratic equation which can be factored to obtain
(x − 1)2 = 0.
Thus, we conclude that x = 1.
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Mathematics Skill Development - Module 5
4. Suppose that a, b and c are real numbers such that ln a = 3, ln b = 2 and ln c = 12. Find the value of
c
ln 2 + ln(ab).
a
Solution:
We use properties of the natural logarithm and the given values of ln a, ln b and ln c to compute
ln
c
+ ln(ab) = ln c − ln(a2 ) + ln a + ln b
a2
= ln c − 2 ln a + ln a + ln b
= ln c − ln a + ln b
= 12 − 3 + 2
= 11.
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