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(a) How long will it lake an investment to double in value if the interest rate is $ 6\% $ compounded continuously? (b) What is the equivalent annual interest rate?

a) $t=11.55$ yearsb) Annual interest $=.06185$ or 6.185$\%$

01:30

Wen Z.

Amrita B.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 8

Exponential Growth and Decay

Derivatives

Differentiation

Campbell University

Oregon State University

University of Nottingham

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Let's find the time it takes to double and investment when compounded continuously, if the interest rate is 6%. So we're going to translate that 2.6 Okay, if the amount doubles than if you start with a not you're going to end with two times a not so two times, two times a not equals a Not each of the 0.0 60 We're solving this for teeth. Let's divide both sides by a not and we have two equals e to the 20.0 60. Now let's take the natural log of both sides, and we have the natural log of two equals 20.0 60. And then we'll divide both sides by 0.6 and we get natural log of two divided by 20.6 and we're gonna proximate that in the calculator and we get about 11.55 years. So that's the amount of time it would take to double under those circumstances. Now, suppose instead of continuous compounding, you have the same amount of time for non continuous annual compounding. Let's find out the interest rate that would be giving us the same amount of money doubling the money, so we want our way to be doubled. We have two times a not, and we're going to use the same amount of time. So we're using approximately 11.55 years. Let's divide both sides of the equation by a not, and we get two equals one plus R to the 11.55 Now to get rid of that power, let's raise both sides to the 1/11 0.55 power, and then we'll subtract one from both sides. So R equals two to the 1/11 20.55 power minus one. And we can put that in the calculator and we get approximately 0.6185 We can convert that back to a percent, and it's approximately 6.185% to get the equivalent amount of money in that amount of time.

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