LESSON X
LESSON 11 EXPONENTIAL AND LOGARITHMIC EQUATIONS
Since any exponential function is one-to-one, then [pic] if and only if [pic].
Examples Solve the following exponential equations. Give exact answers. No decimal approximations.
1. [pic]
Using the one-to-one property: [pic] [pic]
Using logarithms base 3: [pic] [pic]
[pic] [pic]
NOTE: [pic]
Using natural logarithms: [pic] [pic] [pic]
[pic] (using a calculator)
Answer: 4
2. [pic]
3. [pic]
Using the one-to-one property: [pic] [pic]
[pic]
Using logarithms base 2: [pic] [pic]
[pic] [pic]
NOTE: [pic]
Using natural logarithms: [pic] [pic]
[pic] [pic]
[pic] [pic]
(using a calculator)
Answer: [pic]
4. [pic]
5. [pic]
6. [pic]
7. [pic]
Using natural logarithms: [pic] [pic]
[pic] [pic]
NOTE: [pic] and [pic]
Using logarithms base 6: [pic] [pic]
[pic] [pic] NOTE: [pic]
Since your calculator does not have logarithm base 6 key, you would have to do a change of bases to obtain an approximation for [pic]. Since your calculator has a natural logarithm key LN , then we obtain that [pic] =
[pic] using the change of base formula that [pic], where [pic] and [pic]. Or, Since your calculator has a common logarithm key LOG , then we obtain that [pic] = [pic] using the change of base formula that [pic], where [pic] and [pic].
Answer: [pic]
8. [pic]
9. [pic]
10. [pic]
11. [pic]
Examples Solve the following logarithmic equations. Give exact answers. No decimal approximations.
1. [pic]
Using the definition of logarithm ([pic] if and only if [pic] ), we will write the logarithmic equation as an exponential equation:
[pic]
Using square roots to solve the equation [pic], we have that
[pic]
Since the base of a logarithm can not be negative, then the solution of [pic] can not be used. Thus, the only solution of the equation [pic] is [pic].
Answer: [pic]
2. [pic]
Using the definition of logarithm ([pic] if and only if [pic] ), we will write the logarithmic equation as an exponential equation:
[pic]
Using cube roots to solve the equation [pic], we have that
[pic]
Answer: [pic]
3. [pic]
Using the definition of logarithm ([pic] if and only if [pic] ), we will write the logarithmic equation as an exponential equation:
[pic]
We need to check that the number [pic] makes the argument of the logarithm positive since the logarithm of a negative number or zero is undefined. The argument of [pic] is x. Thus, x will be positive when [pic]. Thus, [pic] is a solution of the equation [pic]. Of course, it is the only solution.
Answer: [pic]
4. [pic]
Recall that log is the notation for the common logarithm, and the base of the common logarithm is 10. Using the definition of logarithm ([pic] if and only if [pic] ), we will write the logarithmic equation as an exponential equation:
[pic]
Solving the equation [pic], we have that [pic].
We need to check that the number [pic] makes the argument of the logarithm positive since the logarithm of a negative number or zero is undefined. The argument of [pic] is [pic].
When [pic], we have that [pic] = [pic] = [pic].
Thus, [pic] is a solution of the equation [pic]. Of course, it is the only solution.
Answer: [pic]
5. [pic]
First, we’ll use the property of logarithms that [pic] = [pic] + [pic] in order to write [pic] as [pic]. Thus,
[pic] [pic]
Now, using the definition of logarithm ([pic] if and only if [pic] ), we will write the logarithmic equation [pic] as an exponential equation:
[pic]
Solving the equation [pic], we have that
[pic] [pic] [pic]
[pic] [pic], [pic]
We need to check that the numbers [pic] and 16 make the argument of the logarithms positive since the logarithm of a negative number or zero is undefined. The argument of [pic] is x and the argument of [pic] is [pic].
When [pic], we have that [pic]. Thus, when [pic], we have that [pic]. However, [pic] is undefined. Thus, [pic] is a solution of the equation [pic], but it is not a solution of the equation [pic].
When [pic], we have that [pic] and [pic]. Thus, 16 is a solution of the equation [pic].
Answer: [pic]
6. [pic]
In order to solve this equation, we will use the fact that any logarithm function is one-to-one. Thus, [pic] if and only if [pic].
Thus, by the one-to-one property, we have that
[pic]
Solving the equation [pic], we have that [pic].
We need to check that the number [pic] makes the argument of the logarithms positive since the logarithm of a negative number or zero is undefined. The argument of [pic] is x and the argument of [pic] is [pic].
When [pic], we have that [pic] and [pic]. Thus, [pic] is a solution of the equation [pic].
Answer: [pic]
7. [pic]
In order to solve this equation, we will use the fact that any logarithm function is one-to-one. Thus, [pic] if and only if [pic].
Thus, by the one-to-one property, we have that
[pic]
Solving the equation [pic], we have that [pic].
We need to check that the number [pic] makes the argument of the logarithms positive since the logarithm of a negative number or zero is undefined. The argument of [pic] is t and the argument of [pic] is [pic].
When [pic], we have that [pic]. Thus, when [pic], we have that [pic]. However, [pic] is undefined. Thus, [pic] is a solution of the equation [pic], but it is not a solution of the equation [pic].
Answer: No solution
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