SOLUTION: Can you please help me with this:) 1. Chris invests $10 000 at 7.2% per year, with monthly compounding periods. How long will it take for his investment to grow to $25 000?

Question 1088272: Can you please help me with this:)

1. Chris invests $10 000 at 7.2% per year, with monthly compounding periods. How long will it take for his investment to grow to $25 000?

Found 2 solutions by jim_thompson5910, addingup:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
We're given:
A = 25000 which is the amount we want after some time (t)
P = 10000 which is the initial deposit (aka principal)
r = 7.2% = 0.072 is the interest rate; we'll use the decimal form
n = 12 to indicate monthly compounding, or 12 times a year compounding

The goal is to find t.

Start with the compound interest formula

Plug in all of the given values

Divide both sides by 10000

Flip the equation

Apply logs to both sides

Use one of the log rules to pull down the exponent

Divide both sides by log(1.006)

Get the approximate log values

Divide both sides by 12

It will take about 12.764306 years. If you round to the neaest whole year, then it will take 13 years.

Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
A = Accrued amount (principal + interest, in this case 25000
P = Principal (initial amount, in this case 10000)
r = rate 0.072
n = number of compounding periods (12)
----------------------------------
In the compound interest formula, time is an exponent:
A = P(1+(r/n))^(nt)
In order to solve for an exponent we have to use logarithms.
Take the formula above, do a little algebra on it, and you get:
t = ln(A/P)/n[ln(1+r/n)] = [ln(A)-ln(P)]/n[ln(1+r/n)]
Now just plug in your numbers:
t = [ln(25000/10000)]/[12(ln(1+(0.072/12)))]
t = (ln25000-ln10000)/12(ln(1+0.006))
t = 10.12-9.21/(0.072)
t = 12.64 years

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