Exponent and LogarithmApplications A.) The number ofwords per minute that a stud

Question

Exponent and LogarithmApplications

A.) The number ofwords per minute that a student can type will increase withpractice according to the equation, N = 100[1 – (0.7)t] where N = the number of words per minute after t days ofpractice. Find the number of words (rounded to the nearestword) a student will type per minute after 10 days ofinstruction.

B.) How long (to thenearest hundredth) will it take for $200 to grow to $500 at 6%interest compounded annually?

C.) How long (to thenearest hundredth) will it take for money to double if it isinvested at 6% compounded quarterly?

Explanation / Answer

For (A), you're given an equation that tells you the number ofwords a student will type after t days of practice, and you want toknow how many words they will type after 10 days. So you wantto substitute 10 for t in your equation. For (B) and (C), you need to have a formula for interestrates. If these questions came from a book, or if you'verecently learned about exponentials and logarithms, you may havebeen shown the equation you need, which can be written A = P(1 + r/n)nt where P stands for the initial amount of money you start with, r is the annual interest rate as a decimal, n is the number of times the rate will be calculated each year (thenumber of times it is compounded per year), t is the number of years that will pass, and A is the amount of money that you will have afterward. In (B), we're given the initial amount P = 200, the final amount A= 500, the interest rate r = 0.06, and the number of times it iscompounded per year n = 1, since it is compounded annually. So using our equation, we can write 500 = 200(1 + 0.06/1)1t and can simplify this to 500 = 200(1.06)t We're asked to solve this for the length of time it will take, t(which will be in years). First we can divide both sides by 200 500/200 = (1.06)t But now, we have the t in the exponent, so we need to uselogarithms to bring it down. If we take log of both sides (orln, both work, I'm not sure which you personally may have learnedto write for this, so I'll write log), we get log(500/200) = log(1.06t) and the properties of logarithms let us rewrite the left side aslog(500) - log(200) and let us rewrite the right side ast*log(1.06), so we have log(500)-log(200) = t*log(1.06) Now we can divide both sides by log(1.06) to solve for t [ log(500) - log(200) ] / log(1.06)   =    t Calculating that out give us the answer in years. For (C), we aren't given specific values for A and P, but we aretold that we want the amount in the end to be double the initialamount. That is, we want A = 2*P. We're also given therate r = 0.06 again, and this time the money will be compounded n =4 times per year because it is compounded quarterly. So wecan set up our equation 2*P = P(1 + 0.06/4)4t They've asked us to solve for t, but first we need to get rid ofthe extra variable P. Luckily, we can divide both sides by Pto get 2 = (1 + 0.06/4)4t Which simplifies to 2 = (1.015)4t Now, we can use logarithms to bring the 4t down again and solve forit. log(2) = log(1.0154t) log(2) = 4t*log(1.015) log(2)/log(1.015) = 4t log(2) / ( 4*log(1.015) ) = t

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