Welcome to a lesson on solving exponential equations for time using common logarithms.
As an example, let's say there are four thousand students attending a
local community college, and the number of students is increasing
by five percent each year. When will the student population reach five thousand?
Using this form of an exponential equation, P sub zero is the starting population, so in this case,
P sub zero would be four thousand. R is the growth rate expressed as a decimal, and because
the population is increasing by five percent each year, R would be five percent as a decimal, which
is zero point zero five. The question is, when will the student population reach five thousand,
where P sub n is the population after n years? So P sub n would be five thousand, and the question
is how long it will take to reach this level. Therefore, n, the number of years, is the unknown.
If we set this up, this would give us the equation five thousand equals four thousand times one
plus zero point zero five raised to the power of n. Simplifying one more time gives us this
equation here: five thousand equals four thousand times one point zero five raised to the power of
n. Because the unknown is in the exponent position, we need to talk about logarithms.
In particular, we will use common logarithms to solve this type of exponential equation.
For a very brief review on common logarithms, the common logarithm is log with base ten,
written as we see here, log of x. Log of x undoes the exponential ten raised to the
power of x. This means that log of ten raised to the power of x is equal to just x. Again,
the log undoes the exponential. Likewise, ten raised to the power of log of x equals x. Again,
these undo each other, leaving us with just x.
It is also true that ten raised to the power of b in exponential form is equivalent to
log base ten of b equals a in logarithmic form. Again here, notice how the base is not given,
so we know this is log base ten. We read this as log of x, and it means the logarithm of the
value of x. It is important to note that this is not multiplication. The log by
itself does not mean anything. This is not log times x. For one more example,
ten raised to the power of three equals one thousand is equivalent to log base ten of one
thousand equals three. Notice how I included the base here for illustration purposes, but because
this is base ten, or common log, we normally would not include this base. But to check this,
we can always start with the base. Ten raised to the power of three must equal one thousand.
There is one property of logarithms that we have to review before we can go back
and solve our exponential equation, and this is the exponent property,
sometimes called the power property of logarithms, which tells us that log of
a raised to the power of r is equal to r times log of a. Notice what has happened here is the
exponent of r has been moved to the front, so now we have a product of r and log of a.
The proof of this is given here, but to save some time,
I am not going to walk through this now. You are welcome to pause and review this.
Now that we have reviewed common logs and our exponent property of logs, let us review the
steps on solving exponential equations using logs, and then we will go back to our example.
Step one, we want to isolate the exponential part of the equation.
Step two, we will take the log of both sides of the equation. Step three,
we will use the exponent property of logs to write the exponential with the variable exponent
multiplying the logarithm. Then, step four, we will divide as needed to solve for the variable.
Going back to our example, we want to begin by isolating the exponential part,
which is one point zero five raised to the power of n. So, for our first step,
we will divide both sides by four thousand. This simplifies to one, and on the left,
we have five thousand divided by four thousand, which simplifies to
five-fourths. So we have five-fourths equals one point zero five raised to the power of n.
Now that we have the exponential part isolated,
we will take the log of both sides of the equation. So, we take the log of the right
and the log of the left. Notice on the right side, we can apply the exponent property,
meaning we can take the exponent of n and move it to the front and multiply n by the
logarithm. So now we have log of five-fourths equals n times log of one point zero five.
Now, because we have a product here, we can solve for n by dividing both sides by log of one point
zero five. This simplifies to one, so now we have just n on the right. On the left, we will
use the calculator to approximate this quotient. So, we have log of five-fourths divided by log
of one point zero five, and if we round to two decimal places or the hundredths place value, we
can say n is approximately four point five seven years. This means, in this situation, it will take
approximately four point five seven years for the student population to reach five thousand.
Let us take a look at a second example. Inflation causes things to cost more and
for our money to buy less. Suppose inflation decreases the value of
money by two percent each year. In other words, if you have one dollar this year,
next year it will only be worth ninety-eight cents. How much will one hundred dollars buy
you in fifteen years, and how long until one hundred dollars is worth fifty dollars?
Let us list the given information. The beginning amount is one hundred dollars, and therefore,
P sub zero is equal to one hundred. R, the rate of change, is negative two percent each year,
and therefore R would be negative zero point zero two. For the first part of the question,
they are asking how much one hundred dollars is worth in fifteen years,
and therefore n is equal to fifteen. To answer this first question,
we will not need logarithms, but we will for the second part.
To find the value of one hundred dollars in fifteen years, we want to find P of fifteen. So,
P of fifteen would be equal to P sub zero, or one hundred, times one plus R,
which is zero point zero two, raised to the power of n, and n is fifteen. So,
we have P sub fifteen equals one hundred. Notice how the base here is going to be
one plus zero point zero two, or zero point nine eight, raised to the power of fifteen.
Now, we will use the calculator. So, we have one hundred times our base of
zero point nine eight raised to the power of fifteen. One hundred dollars would be
worth approximately seventy-three dollars and eighty-six cents in fifteen years.
Now, we will have to use a second setup to determine how long until one hundred dollars is
worth fifty dollars. To answer the second question, P sub zero is still one hundred,
R is still negative zero point zero two, but now we do not know n because we are trying to
find when P sub n, the value after n years, is equal to fifty dollars. So now, the setup
is going to be the equation fifty equals P sub zero, which is one hundred. We already know that
one plus R is going to be zero point nine eight from the last example, raised to the power of n.
Notice here, because the variable is in the exponent position,
we will have to use logarithms to solve this. So again, the first step is to isolate the
exponential part. We will divide both sides by one hundred. This simplifies to one-half,
so we have one-half equals zero point nine eight raised to the power of n.
Now, we will take the log of both sides of the equation. By taking the log of both
sides of the equation, we can now apply the power property, or exponent property,
of logarithms and take the exponent of n and move it to the front and multiply it by the
logarithm. So now we have log of one-half equals n times log of zero point nine eight.
To solve for n, because this is a product, we can just divide by log of zero point nine eight. This
simplifies to one. So, we have n is going to be approximately equal to this quotient here.
Let us round to the nearest hundredth. So, we will go back to the calculator. We have log
of one-half divided by log of zero point nine eight. If we round to the nearest hundredth,
the answer would be approximately thirty-four point thirty-one years.
I hope you found this helpful.