Module 6: Solving Exponential Equations



Section II: Exponential and Logarithmic Functions

[pic]

Module 6: Solving Exponential Equations and More

[pic] EXAMPLE: Solve the equation [pic] for x. Obtain an exact solution.

SOLUTION:

This equation is so easy to solve that most of us can do it in our heads, without showing any work. (Clearly, [pic] since [pic].) But it might be helpful to notice what “work” would have helped us solve this equation, so that we can learn how to solve exponential equations that we can’t do in our heads.

If it is possible, the easiest way to solve exponential equations is to write both sides of the equation as a power of the same base:

[pic]

Since exponential functions are one-to-one, we can now assume that since both sides of the equation are expressed as powers of the same base, the exponents must be equal, which allows us to conclude that [pic]

But this technique requires us to be able to write both sides of the equation in terms of the same base. In order to learn how to solve exponential equations involving expressions that don’t allow us to do this so easily, let’s investigate another method of solving exponential equations. Recall that the log-of-powers law allows us to move powers out of the exponent. Thus, let’s utilize a logarithmic function to bring the x out of the exponent:

[pic]

[pic]

[pic] EXAMPLE: Solve the equation [pic] for x. Obtain an exact solution.

SOLUTION:

Since both 8 and 16 can be expressed as powers of the same base (2) we can solve the equation using the “same base method.”

[pic]

But, again, we are lucky to be able to use this method, since both sides of the equation can be expressed as powers of the same base. Another way to solve this equation is to use logarithms. We use the natural logarithm in this case since it is easiest to write (only two letters!) and it is the usually easiest logarithm to estimate on calculators.

[pic]

[pic]

In general, when solving exponential equations we are not usually able to use the “same base method” but we can ALWAYS use the method involving logarithms. The next few examples we NEED logarithms to solve the equations. We will use the natural logarithm here for the same reasons given in the example above. Although ANY logarithm can be used, the natural logarithm (or common logarithm) is recommended.

[pic] EXAMPLE: Solve the equation [pic] for x. Obtain an exact solution.

SOLUTION:

[pic]

Although our solution isn’t very user friendly (most of us don’t have much intuition about the number [pic]), we were asked to find and exact solution, and we can’t express this number in a simpler way. So we need to leave this as our solution. If you want to approximate it, that is fine, but the approximation must be given in addition to the exact solution, NOT instead of the exact solution.

[pic]

[pic] EXAMPLE: Solve the equation [pic] for p. Obtain an exact solution.

SOLUTION:

[pic]

[pic] EXAMPLE: Solve the equation [pic] for t. Obtain an exact solution.

SOLUTION:

[pic]

[pic]

[pic] EXAMPLE: Solve the equation [pic] for x. Obtain an exact solution.

SOLUTION:

[pic][pic]

[pic] EXAMPLE: Solve the equation [pic] for x. Obtain an exact solution.

|[pic] |CLICK HERE FOR THE SOLUTION |

[pic]

[pic] EXAMPLE: Solve the equation [pic] for x. Obtain an exact solution.

|[pic] |CLICK HERE FOR THE SOLUTION |

[pic]

[pic] EXAMPLE: Solve the equation [pic] for x. Obtain an exact solution.

|[pic] |CLICK HERE FOR THE SOLUTION |

[pic]

[pic] EXAMPLE: If [pic], find [pic].

|[pic] |CLICK HERE FOR THE SOLUTION |

[pic]

[pic] EXAMPLE: If [pic], find [pic].

|[pic] |CLICK HERE FOR THE SOLUTION |

[pic] EXAMPLE: If [pic], find [pic].

|[pic] |CLICK HERE FOR THE SOLUTION |

[pic]

[pic] EXAMPLE: If [pic], find [pic].

|[pic] |CLICK HERE FOR THE SOLUTION |

[pic]

APPLICATIONS INVOVLVING EXPONENTIAL FUNCTIONS

[pic] EXAMPLE: Suppose that a population grows according to the model [pic] where t represents years after January 1, 2002.

a. What is the doubling-time for the population?

b. How long will it take for the population to increase 10%?

SOLUTIONS:

a. In order to find the doubling-time, we need to determine how long it takes for the population to double. Since the population has initial size A, we need to find t such that [pic]. (Although we don’t know what number A represents, we can still determine when the population is 2A.)

[pic]

Therefore, the doubling-time is about 9.9 years. (One thing worth keeping in mind about the doubling-time is that it represents the time it takes for the population to double no matter when you start counting. So it takes the same 9.9 years for the population to go from 2A to 4A or from 5A to 10A.)

b. If it grows 10%. then the population will be [pic], so we need to find the time t such that [pic]:

[pic]

Therefore, it takes about 1.36 years for the population to grow by 10%.

[pic]

[pic] Try this one yourself and check your answer.

Iodine-131 was one of the radioactive substances released into the atmosphere during the Chernobyl disaster in the former Soviet Union in 1987. If iodine-131 decays at the continuous daily rate of 8.66%, find its half-life.

SOLUTION:

If you have A grams of iodine-131 then the amount left after t days is given by the function

[pic]

To find the half-life, we need to determine how long it takes for the amount of iodine to shrink by 50%, so we need to solve [pic] for t.

[pic]

So the half-life of iodine-131 is about 8 days.

[pic] EXAMPLE: The half-life of radium-226 is 1620 years. What is the continuous annual rate of decay for radium-226?

SOLUTION:

In order to answer the question, we need to solve the equation below for k.

[pic]

So the continuous annual decay rate for radium-226 is about 0.043%.

[pic]

[pic] EXAMPLE: Suppose that the population of the island-nation Enlargia is currently growing at the simple annual rate of 3.3%. At what continuous annual rate is the population growing?

SOLUTION:

Since the population of Enlargia is growing at the simple annual rate of 3.3%, the function [pic] models the population t years from now, where A is the current population. If we want the continuous annual growth rate, we will need to find k so that [pic]. Since we want both

[pic] and [pic]

to represent the Enlargia’s population, we need them to be equivalent:

[pic]

Therefore, Enlargia’s 3.3% simple annual growth rate is approximately equivalent to a 3.25% continuous annual growth rate. Notice that the continuous annual rate is smaller than the simple annual rate since the continuous rate allows “interest to earn interest.”

[pic] Try this one yourself and check your answer.

The population of the island-nation of Hypothetica is decreasing at the simple annual rate of 4%. At what continuous annual rate is the population of Hypothetica decreasing?

SOLUTION:

To answer the question we need to fink k so that [pic]:

[pic]

Therefore, Hypothetica’s 4% simple annual decay rate is approximately equivalent to a 4.08% continuous annual decay rate.

[pic]

[pic] EXAMPLE: Bonny and Sunny deposit $800 simultaneously at two different banks.

a. If Bonny’s money earns 5% simple annual interest, find a rule for the function b that gives the amount in Bonny’s account t years after making the deposit.

b. At Sunny’s bank, interest is compounded continuously. If Sunny earns the same annual yield as does Bonny (i.e., at the end of each year, their accounts have the same balance – which implies that their accounts always have the same balance), what continuous annual interest rate does Sunny’s bank give?

SOLUTION:

a. The desired function is [pic].

b. Since Sunny also deposits $800, if her continuous interest rate is k then the value of her account is given by the function

[pic]

Since Sunny’s annual yield is the same as Bonny’s, we know that [pic]. We can use this fact to find k:

[pic]

So Sunny’s bank gives about 4.88% continuous annual interest.

[pic]

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