George invests $ 15 000 at 7.2%/a compounded monthly . How long will it take for his investment to grow to $ 34 000
let the number of months be n
i = .072/12 = .006
15000(1.006)^n = 34000
1.006^n = 2.266666...
log 1.006^n = log 2.66666..
n log 1.006 = log 2.26666...
n = log 2.266666.../log 1.006
= 136.79 or 137 months
= 11 years and 5 months
To find out how long it will take for George's investment to grow to $34,000, we need to use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment ($34,000 in this case)
P = the principal amount (initial investment of $15,000)
r = annual interest rate as a decimal (7.2% = 0.072)
n = number of times interest is compounded per year (monthly compounding means n = 12)
t = time in years
Now, let's substitute the known values into the formula and solve for t:
34,000 = 15,000(1 + 0.072/12)^(12t)
Dividing both sides of the equation by 15,000:
34,000/15,000 = (1 + 0.072/12)^(12t)
Simplifying the fraction on the right side:
34/15 = (1 + 0.006)^12t
Now we need to isolate the exponent:
34/15 = (1.006)^12t
Taking the natural logarithm (ln) of both sides:
ln(34/15) = ln(1.006)^12t
Using the property of logarithms allowing us to move the exponent down:
ln(34/15) = 12t * ln(1.006)
Now we can calculate t by dividing both sides by 12 times ln(1.006):
t = ln(34/15) / (12 * ln(1.006))
Using a calculator, evaluate the right side to find the value of t, which will give the number of years it will take for the investment to grow to $34,000.