# Ex: Exponential Function Applications – Increasing Investment Value (Change of Base Used)

In 2011 TabaSue had a \$2,000 savings bond that had reached maturity
and she invested \$6,000 in a savings account that paid 4.1% interest compounded annually. The formula that can be used to represent the growth of TabaSue’s
savings can be modeled by A of t equals 2000 plus 6000 times 1.041 raised to the power of t where
A of t is the accrued value of the investments t years after 2011. So it’s important to
note here that t is the number of years after 2011, t is not the actual year. We want to use the information to answer the following two questions. First determine the
accrued value of TabaSue’s investments in the year 2020. Round your answer to two decimal places. So the first step is to determine what value of t represents 2020. Well we know t equals zero. Represents the year 2011. And since 2020 is nine years after 2011, t equals nine represents the year 2020. If we ever have a difficult time determining the value of t, t is always going to be
equal to the desired year, in this case 2020, minus the base year, in this case 2011, which gives us a t value of nine. Now that we know t equals
nine represents the year 2020, to find the accrued value, we want to determine A of nine. So we substitute nine for t, so we’d have 2000 plus 6000 times 1.041, raised to the power of nine. And now we’ll go to the calculator to approximate this value
to two decimal places. So 2000 plus 6000 times 1.041, I’m going to raise this
to the ninth power. And press enter. So the approximate value
would be \$10,614.06. Now we want to determine
in what year TabaSue’s investment will reach an
accrued value of \$12,530.80. And we’re asked to round
to the nearest year. So now we’re given the value of A of t, A of t equals \$12,530.80. We want to determine the
value of t or the input That would give us this
output of this function value. So we’ll substitute this value for A of t, creating an exponential equation which we’ll then solve for t. Meaning we want to solve the equation 12,530.8 equals 2,000 plus 6,000 times 1.041, raised to the power of t. Now when we solve this equation for t, this will give us the
number of years after 2011 so to find the actual year, we’ll have to add the value
of t to the base year of 2011. Let’s go ahead and solve
this on the next slide. We first want to isolate the exponential part of the equation. Meaning we want to isolate 1.041 raised to the power of t. So we’ll begin by subtracting 2,000 on both sides of the equation. So this will give us 10,530.8 equals, this is zero, 6,000 times 1.041 raised to the power of t. And now for the next step we’ll
divide both sides by 6000. This simplifies to one. Now what we don’t want
to do here on the left is find this quotient
and round this value. And then perform another
calculation and round again. Because that would give
us too much of an error. So we’re going to go ahead and leave the left side as this quotient, so we have 10,530.8 divided by 6,000 equals 1.041 raised to the power of t. Now that we’ve isolated
the exponential part of the equation we have a choice on how to solve this equation. We could apply the power property of logarithms that was given here, and then solve for t, or we can write the exponential
equation as a log equation and then apply the change of base formula. For this example we’ll write
the exponential equation as the log equation and then use the change of base formula. To do this we need to be
able to identify the base, the exponent, and the number. So notice for our exponential equation, the base is 1.041, the exponent is t, and the number would be this quotient. Which means as a log equation, we would have log base 1.041 of the quotient 10,530.8 divided by 6,000
must equal the exponent t. And now to evaluate this logarithm, we’ll use the change of
base formula given here and we’ll use the common log. And therefore t is equal to the common log of the quotient 10,530.8 divided by 6,000 divided by the common
log of the base 1.041. And now we’ll go to the calculator and round t to the nearest year. The common log key is here. So common log of the quotient 10,530.8 divided by 6,000. Closed parenthesis. Divided by the common log of 1.041, closed parenthesis and enter. We’re asked to round to the nearest year. So we’ll round this to 14 years. Now that we know that t
is approximately 14 years, we can now determine in what year the investment account will reach
the value of \$12,530.80. The year would be equal to 2011 plus t or in this case 2011 plus 14, which gives us 2025. I hope you found this helpful.