Transcript Slide 1

Exponential and Logarithmic
Functions
Copyright © Cengage Learning. All rights reserved.
Exponential and
4.5
Logarithmic Equations
Copyright © Cengage Learning. All rights reserved.
Objectives
► Exponential Equations
► Logarithmic Equations
► Compound Interest
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Exponential Equations
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Exponential Equations
An exponential equation is one in which the variable occurs
in the exponent. For example,
2x = 7
The variable x presents a difficulty because it is in the
exponent.
To deal with this difficulty, we take the logarithm of each
side and then use the Laws of Logarithms to “bring down x”
from the exponent.
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Exponential Equations
2x = 7
ln 2x = ln 7
x ln 2 = ln 7
Given Equation
Take In of each side
Law 3 (Bring down exponent)
Solve for x
 2.807
Calculator
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Exponential Equations
Recall that Law 3 of the Laws of Logarithms says that
loga AC = C loga A.
The method that we used to solve 2x = 7 is typical of how
we solve exponential equations in general.
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Example 1 – Solving an Exponential Equation
Find the solution of the equation 3x + 2 = 7, rounded to six
decimal places.
Solution:
We take the common logarithm of each side and use
Law 3.
3x + 2 = 7
log(3x + 2) = log 7
Given Equation
Take log of each side
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Example 1 – Solution
(x + 2)log 3 = log 7
x+2=
cont’d
Law 3 (bring down exponent)
Divide by log 3
Subtract 2
 –0.228756
Calculator
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Example 1 – Solution
cont’d
Check Your Answer
Substituting x = –0.228756 into the original equation and
using a calculator, we get
3(–0.228756) + 2  7
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Example 4 – An Exponential Equation of Quadratic Type
Solve the equation e2x – ex – 6 = 0.
Solution:
To isolate the exponential term, we factor.
e2x – ex – 6 = 0
(ex)2 – ex – 6 = 0
Given Equation
Law of Exponents
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Example 4 – Solution
(ex – 3)(ex + 2) = 0
ex – 3 = 0
ex = 3
or
cont’d
Factor (a quadratic in ex)
ex + 2 = 0
Zero-Product Property
ex = –2
The equation ex = 3 leads to x = ln 3.
But the equation ex = –2 has no solution because ex > 0 for
all x.
Thus, x = ln 3  1.0986 is the only solution.
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Logarithmic Equations
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Logarithmic Equations
A logarithmic equation is one in which a logarithm of the
variable occurs.
For example,
log2(x + 2) = 5
To solve for x, we write the equation in exponential form.
x + 2 = 25
x = 32 – 2 = 30
Exponential form
Solve for x
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Logarithmic Equations
Another way of looking at the first step is to raise the base,
2, to each side of the equation.
2log2(x + 2) = 25
x + 2 = 25
x = 32 – 2 = 30
Raise 2 to each side
Property of logarithms
Solve for x
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Logarithmic Equations
The method used to solve this simple problem is typical.
We summarize the steps as follows.
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Example 6 – Solving Logarithmic Equations
Solve each equation for x.
(a) ln x = 8
(b) log2(25 – x) = 3
Solution:
(a)
ln x = 8
x = e8
Given equation
Exponential form
Therefore, x = e8  2981.
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Example 6 – Solution
cont’d
We can also solve this problem another way:
ln x = 8
eln x = e8
x = e8
Given equation
Raise e to each side
Property of ln
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Example 6 – Solution
cont’d
(b) The first step is to rewrite the equation in exponential
form.
log2(25 – x) = 3
25 – x = 23
Given equation
Exponential form
(or raise 2 to each side)
25 – x = 8
x = 25 – 8 = 17
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Example 6 – Solution
cont’d
Check Your Answer
If x = 17, we get
log2(25 – 17) = log2 8 = 3
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Logarithmic Equations
Logarithmic equations are used in determining the amount
of light that reaches various depths in a lake. (This
information helps biologists to determine the types of life a
lake can support.)
As light passes through water (or other transparent
materials such as glass or plastic), some of the light is
absorbed.
It’s easy to see that the murkier the water, the more light is
absorbed. The exact relationship between light absorption
and the distance light travels in a material is described in
the next example.
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Example 10 – Transparency of a Lake
If I0 and I denote the intensity of light before and after going
through a material and x is the distance (in feet) the light
travels in the material, then according to the
Beer Lambert Law,
where k is a constant depending on the type of material.
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Example 10 – Transparency of a Lake cont’d
(a) Solve the equation for I.
(b) For a certain lake k = 0.025, and the light intensity is
I0 = 14 lumens (lm).
Find the light intensity at a depth of 20 ft.
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Example 10 – Solution
(a) We first isolate the logarithmic term.
Given equation
Multiply by –k
Exponential form
I = I0e–kx
Multiply by I0
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Example 10 – Solution
cont’d
(b) We find I using the formula from part (a).
I = I0e–kx
From part (a)
= 14e(–0.025)(20)
I0 = 14, k = 0.025, x = 20
 8.49
Calculator
The light intensity at a depth of 20 ft is about 8.5 lm.
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Compound Interest
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Compound Interest
If a principal P is invested at an interest rate r for a period
of t years, then the amount A of the investment is given by
A = P(1 + r)
Simple interest (for one year)
Interest compounded n times per year
A(t) = Pert
Interest compounded continuously
We can use logarithms to determine the time it takes for the
principal to increase to a given amount.
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Example 11 – Finding the Term for an Investment to Double
A sum of $5000 is invested at an interest rate of 5% per
year.
Find the time required for the money to double if the
interest is compounded according to the following method.
(a) Semiannually
(b) Continuously
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Example 11 – Solution
(a) We use the formula for compound interest with
P = $5000, A(t) = $10,000, r = 0.05, and n = 2 and solve
the resulting exponential equation for t.
(1.025)2t = 2
log 1.0252t = log 2
2t log 1.025 = log 2
Divide by 5000
Take log of each side
Law 3 (bring down the
exponent)
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Example 11 – Solution
cont’d
Divide by 2 log 1.025
t  14.04
Calculator
The money will double in 14.04 years.
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Example 11 – Solution
cont’d
(b) We use the formula for continuously compounded
interest with P = $5000, A(t) = $10,000, and r = 0.05 and
solve the resulting exponential equation for t.
5000e0.05t = 10,000
e0.05t = 2
ln e0.05t = ln 2
0.05t = ln 2
Pert = A
Divide by 5000
Take ln of each side
Property of ln
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Example 11 – Solution
cont’d
Divide by 0.05
t  13.86
Calculator
The money will double in 13.86 years.
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