Approximately how long will it take $1000 to grow to $5006 at 8% annual interest and compounded 12 times p/yr?

Question

12 times per year?

(a) 18.1 years
(b) 25.7 years
(c) 20.2 years
(d) 32.8 years
Any help will be appreciated or maybe some explanation would also help...Thanx

Answer 1

Hi,

The correct answer is C. the formula you are using is:

A = P(1 + r/n)^(n*y) where A will be the amount of money you end up with, P ids the principal or starting amount of money, r is the percentage rate as a decimal, n is the number of times interest is paid each year, and y is the number of years.

With your numbers, this equation becomes:

5006 = 1000(1 + .08/12)^(12y)

That simplifies to 5.006 = (1.006666666)^(12y)
Since your variable is part of the exponent, take the log of both sides. log 5.006 = log(1.006666666)^(12y)
When you have the log of a number raised to a power, you can move the exponent out in front of the log and multiply by it instead. This makes the equation: log 5.006 = (12y) * log(1.006666666)
To get y all alone, divide both sides of the equation by 12 * log(1.006666666), so
log5.006/(12 * log(1.006666666) = (12y) * log(1.006666666)/(12 * log(1.006666666).

This simplifies to:log5.006/(12 * log(1.006666666) = y
and 20.2 = y

I hope this helps.

Answer 2

5006 = 1000(1+ (0.08/12))^(time)
5.006 = 1.0066667^time
log5.006 / log1.0066667 = time
time = 20.2 years or 242 months

Answer 3

Let x be the number of years required.

By the definition of compound interest,
1000(1+0.08/12)^(12x) = 5006

Solve for x,
x = [ ln(5.006)/ln(1+0.08/12)]/12 = 20.2 years.

Answer 4

P = P1(1+r/12)^12y
5006 = 1000(1+.08/12)^12y
5006/1000 = 1.006667^12y
5.006 = 1.006667^12y
ln(5.006) = 12y ln(1.006667)
ln(5.006)/ln(1.006667 =12y
y = ln (5.006)/(12* ln(1.006667)
y= 1.61064/(12*0.006645)
y = 20.2 years

Answer 5

P=1000$
A=5006$
n=?
r=8%
r per month=.75
5006=1000[1+.0075]^n
1.0075^n=5.006
n*log 1.0075=log5.006
n=log5.006/log1.0075
=.6994/.00324
=215.86 months
=17.98
approx18 years
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