# If an amount of money is invested for 29 years with an interest rate of approximately 10%, the...

## Question:

If an amount of money is invested for 29 years with an interest rate of approximately 10%, the value will quadruple two times.

If \$P is invested, write an exponential expression that represents the increase on that investment.

## Exponential Growth

Many things in the world can be well-modeled by exponential growth. If a quantity grows exponentially, then when we add a certain amount of time, the quantity multiplies by a certain factor. For example, a population of frogs might double every {eq}t {/eq} months. A mathematical model for this situation might look like this:

$$\mathrm{P=P_0\times 2^{\frac{m}{t}}}$$

Here, {eq}P {/eq} represents the population of the frogs after {eq}m {/eq} months. The quantity {eq}\mathrm{P_0} {/eq} represents the population at the start of the model time. We can see the multiplying factor, {eq}t {/eq} as the base of the exponential function. Each time {eq}m {/eq} increases by {eq}t {/eq}, the exponent will increase by 1 and the population will double.

An interest rate of {eq}10\% {/eq} means that the investment will be multiplied by {eq}1.1 {/eq} each year. We can model this situation like this:

$$A=P\times \left(1.1\right)^t$$

Here, {eq}A {/eq} represents the total amount after {eq}t {/eq} years of interest accrues. {eq}P {/eq} is the original investment amount. So in this problem, we are told that a certain amount of money is invested for {eq}29\ \mathrm{years} {/eq}. Using our model, we find that the total amount in the account would be:

$$A=P\times \left(1.1\right)^{29}$$

In fact, we can check to see what value {eq}\mathrm{1.1^{29}} {/eq} has. Using a calculator, we see that it is nearly {eq}16 {/eq}. That is what we would multiply the original amount by if it were to quadruple twice. The value is not exactly {eq}16 {/eq} because the interest rate is only approximately {eq}10\% {/eq}. We are not asked to find the exact rate, so we will proceed with this approximate model.

Now, we are asked to make an exponential expression that represents the increase on the original amount. That means we need to subtract the final value minus the initial value.

\begin{align*} A-P&=P\times \left(1.1\right)^{29}-P\\ A-P&=P\left(1.1^{29}-1\right) \end{align*}

So there's our expression: {eq}\mathrm{P\left(1.1^{29}-1\right)} {/eq}