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Find the rate of interest earned by an investment that grows exponentially, if a $\$ 1200 dollars investment increases to $\$ 2700$ in $9^{1 / 2}$ years.

$$8.54 \%$$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 7

Applications of Exponential and Logarithmic Functions

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There's a $1,200 investment which grows to $2700 in 9.5 years. What is its rate of interest? So we're gonna be using the function you see bubbled in and green here. And let's start by just plugging in the information that we do have. So we see that a. So what it grew to be would be that $2700 Which is equal to the principal amount of $1,200 times e. The power of our rates. That's what we're trying to find times T. Which is our time of 9.5 years. So 9.5 Let's go ahead and start to simplify this. Let's divide both sides by that 1200 and we get 2.25 on the left hand side Which is equal to E. to our times 9.5. Now to go ahead and get rid of that exponent. Let's take the natural log of both sides. So Ellen of 2.25 Gives us 0.81, Which is then equal to our times 9.5. Because by taking the natural log of this side allows us to get rid of E and L. Links they cancel. That will bring that exponent down. Now we just need to divide both sides by 9.5 and we see that ours then equal to 0.085. So if you multiply that by 100 we get a rate of interest which is equal to 8.5%.

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