1. Exponential and Logarithmic Equations
Section 11.7
Exponential and Logarithmic Equations

2. Objectives Solve exponential equations Solve logarithmic equations
Use exponential and logarithmic equations to solve application problems

3. Objective 1: Solve Exponential Equations
An exponential equation contains a variable in one of its exponents. Some examples of exponential equations are

4. Objective 1: Solve Exponential Equations
Exponent Property of Equality: If two exponential expressions with the same base are equal, their exponents are equal. For any real number b, where b ≠ –1, 0, or 1,
bx = by is equivalent to x = y
Logarithm Property of Equality: If two positive numbers are equal, the logarithms base-b of the numbers are equal. For any positive number b, where b ≠ 1, and positive numbers x and y,
logb x = logb y is equivalent to x = y

5. Objective 1: Solve Exponential Equations
Strategy for Solving Exponential Equations:
1. Isolate one of the exponential expressions in the equation.
2. If both sides of the equation can be written as exponential expressions with the same base, do so. Then set the exponents equal and solve the resulting equation.
3. If step 2 is difficult or impossible, take the common or natural logarithm on both sides. Use the power rule of logarithms to write the variable exponent as a factor, and then solve the resulting equation.
4. Check the results in the original equation.

6. EXAMPLE 1
Solve: 3x + 1 = 81
Strategy We will express the right side of the equation as a power of 3.
Why If each side of the equation is expressed as a power of the same base (in this case, 3), we can use the exponent property of equality to set the exponents equal and solve for x.

7. EXAMPLE 1 Solution Solve: 3x + 1 = 81
Solve for x by subtracting 1 from both sides.
The solution is 3 and the solution set is {3}. To check this result, we substitute 3 for x in the original equation.

8. EXAMPLE 4
Solve: 6x − 3 = 2x
Strategy We will take the common logarithm of both sides of the equation.
Why We can then use the power rule of logarithms to move the expressions x – 3 and x from their current positions as exponents to positions as factors.

9. EXAMPLE 4 Solution Solve: 6x − 3 = 2x
The log of a power is the power times the log. The expression(x – 3) is now a factor of(x – 3) log 6
and not an exponent.
To get the terms involving x on the left side, add 3 log 6 and x log 2 subtract on both sides.
This is the exact solution.
Approximate. On a reverse-entry scientific calculator,
press: 3  6 LOG  ( 6 LOG  2 LOG ) =.
To four decimal places, the solution is To check the approximate solution, we substitute for each x in 6x − 3 = 2x. The resulting values on the left and right sides of the equation should be approximately equal.

10. Objective 2: Solve Logarithmic Equations
A logarithmic equation is an equation with a logarithmic expression that contains a variable. Some examples of logarithmic equations are

11. EXAMPLE 6
Solve: log 5x = 3
Strategy Recall that log 5x = log10 5x. To solve log 5x = 3, we will instead write and solve an equivalent base-10 exponential equation.
Why The resulting exponential equation is easier to solve because the variable term is isolated on one side.

12. EXAMPLE 6 Solution Solve: log 5x = 3 log10 5x = 3
The base of the logarithm is 10.
The solution is 200 and the solution set is {200}.

13. Objective 3: Use Exponential and Logarithmic Equations to Solve Application Problems
Experiments have determined the time it takes for half of a sample of a radioactive material to decompose. This time is a constant, called the material’s half-life.
Radioactive Decay Formula If A is the amount of radioactive material present at time t, A0 was the amount present at t = 0, and h is the material’s half-life, then
A = A02−t/h
When living organisms die, the oxygen–carbon dioxide cycle common to all living things ceases, and carbon-14, a radioactive isotope with a half-life of 5,700 years, is no longer absorbed. By measuring the amount of carbon-14 present in an ancient object, archaeologists can estimate the object’s age by using the radioactive decay formula.

14. Objective 3: Use Exponential and Logarithmic Equations to Solve Application Problems
Exponential Growth Formula If P is the population at some time t, P0 is the initial population at t = 0, and k depends on the rate of growth, then
P = P0ekt

15. EXAMPLE 11
Carbon-14 Dating. How old is a piece of wood that retains only one-third of its original carbon-14 content?
Strategy If A0 is the original carbon-14 content, then today’s content A = We will substitute for A and 5,700 for h in the radioactive decay formula and solve for t.
Why The value of t is the estimated age of the piece of wood.

16. EXAMPLE 11
Carbon-14 Dating. How old is a piece of wood that retains only one-third of its original carbon-14 content?
Solution
Approximate. On a reverse-entry scientific calculator,
press: 5700  2 LOG  2 LOG ) =.
The piece of wood is approximately 9,000 years old.

17. EXAMPLE 12
Population Growth. The bacteria in a laboratory culture increased from an initial population of 500 to 1,500 in 3 hours. How long will it take for the population to reach 10,000?
Strategy We will substitute 500 for P0, 1,500 for P, and 3 for t into the exponential growth model and solve for k.
Why Once we know the value of k, we can substitute 10,000 for P, 500 for P0, and the value of k into the exponential growth model and solve for the time t.

18. EXAMPLE 12
Population Growth. The bacteria in a laboratory culture increased from an initial population of 500 to 1,500 in 3 hours. How long will it take for the population to reach 10,000?
Solution

19. EXAMPLE 12
Population Growth. The bacteria in a laboratory culture increased from an initial population of 500 to 1,500 in 3 hours. How long will it take for the population to reach 10,000?
Solution
To find when the population will reach 10,000, we substitute 10,000 for P, 500 for P0, and for k in the growth model and solve for t :
The culture will reach 10,000 bacteria in about 8 hours.
Approximate. On a reverse-entry scientific calculator, press:
3  20 LN  3 LN ) =.