Question Video: Solving Exponential Equations with Continuously Compounded Interest | Nagwa Question Video: Solving Exponential Equations with Continuously Compounded Interest | Nagwa

# Question Video: Solving Exponential Equations with Continuously Compounded Interest Mathematics • Second Year of Secondary School

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Chloe invested \$3000 at an interest rate of 2% compounded continuously. After how many years and months will her investment be worth \$3500?

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### Video Transcript

Chloe invested 3000 dollars at an interest rate of two percent compounded continuously. After how many years and months will her investment be worth 3500 dollars?

In this question, we’re told that Chloe has a starting amount of 3000 dollars, and it’s invested at an interest rate of two percent. But this interest is compounded continuously. Continuous compounding is a mathematical limit that compound interest can reach if it is calculated and reinvested into an account’s balance over a theoretically indefinite number of periods. If that all sounds very complicated, don’t worry there’s a formula to help.

This formula tells us that for the variables of 𝑉, which represents the value at time 𝑡; 𝑃, the principal or starting amount; 𝑟, which is the annual interest rate; and 𝑡, which is the number of years, then 𝑉 equals 𝑃 times 𝑒 to the power of 𝑟𝑡. Notice that this formula makes use of the mathematical constant 𝑒. Let’s see how we can apply this formula in the context of the question.

As we’re told that Chloe has 3500 dollars after a certain time period, then that’s what we fill in for our variable 𝑉. The starting amount was 3000 dollars. The constant 𝑒 remains the same. The interest rate is two percent, which we write as the decimal 0.02. Finally, the number of years is what we wish to calculate. So, we can leave that with the variable 𝑡. We then take this equation 3500 equals 3000 multiplied by 𝑒 to the power of 0.02𝑡 and rearrange it to solve for 𝑡.

The first thing we would do is to divide both sides by 3000, and then we can simplify the fraction on the left-hand side. 3500 divided by 3000 is equivalent to seven-sixths. In order to extract the value of 𝑡, which is part of the exponent of our constant 𝑒, we’ll need to remember some of our work with logarithms. Because the natural log of 𝑒 to the power of 𝑥 is equal to 𝑥, then that means to solve this equation, we’ll need to find the natural log of both sides. Because the natural log of 𝑒 to the power of 𝑥 is equal to 𝑥, that means when we take the natural log of 𝑒 to the power of 0.02𝑡, we simply get the value of 0.02𝑡.

Finally, to solve for 𝑡, we need to divide by 0.02. That means that 𝑡 is equal to the natural log of seven-sixths divided by 0.02. Typing this calculation into our calculator gives us that 𝑡 is equal to 7.707 and so on. We should remember that the unit for 𝑡 is years. However, this isn’t sufficient for the answer as we were asked to give it in terms of years and months. We know that there will be seven years. But how many months are there in 0.707 and so on years? Well, if we take this decimal value and multiply it by 12, it gives us 8.49 and so on months. We now need to give this to the appropriate number of months.

Remember that we’re talking about this investment problem. Chloe started with 3000 dollars. And we need to work out when her investment will be worth 3500 dollars. This means that if we rounded down the number of months to give an answer of seven years and eight months, it would be incorrect since Chloe didn’t actually get the amount of 3500 dollars until seven years and 8.49 and so on months. Therefore, we need to write the number of months up to give seven years and nine months. And so, that’s our answer in terms of years and months for when Chloe’s investment was worth 3500 dollars.

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