Kelly is planning to renovate her house. She intends to spend $200000. She invests $150000 in an account at 5.3% interest, compounded monthly. It will take Kelly ____ years to meet her goal, to the nearest tenth of a year.
P = Po*(1+r)^n
r = (5.3%/12)/100% = 0.004417 = Monthly
% rate expressed as a decimal.
n = 12Comp/yr * T yrs = 12t Compounding
periods.
P = 150,000(1.004417)^12t = 200.000
(1.004417)^12t = 200,000/150,000=1.33333
12t*Log(1.004417) = Log1.33333
12t*1.914*10^-3 = 124.9377*10^-3
12t = 65.28
t = 5.44 Yrs.
Thanks Henry!
Glad I could help!
To determine the number of years it will take for Kelly to meet her renovation goal, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount (Kelly's goal)
P = the principal amount (Kelly's investment)
r = annual interest rate (in decimal form)
n = number of times interest is compounded per year
t = number of years
In this case, Kelly's goal is $200,000. Her principal investment is $150,000. The annual interest rate is 5.3% or 0.053 (in decimal form). The interest is compounded monthly, so n = 12 (12 months in a year).
Let's plug in these values into the formula and solve for t:
$200,000 = $150,000(1 + 0.053/12)^(12t)
Divide both sides of the equation by $150,000:
$200,000/$150,000 = (1 + 0.053/12)^(12t)
1.33333... = (1.0044167)^(12t)
To solve for t, we need to take the logarithm of both sides:
log(1.33333...) = log((1.0044167)^(12t))
Using logarithm properties, we can bring the exponent down:
log(1.33333...) = 12t * log(1.0044167)
Now, divide both sides by 12 * log(1.0044167):
t = log(1.33333...)/(12 * log(1.0044167))
Using a calculator, we can find that log(1.33333...) ≈ 0.12385, and log(1.0044167) ≈ 0.00029319.
Substituting these values into the equation:
t ≈ 0.12385 / (12 * 0.00029319)
t ≈ 0.12385 / 0.00351828
t ≈ 35.24
Rounding to the nearest tenth of a year, it will take Kelly approximately 35.2 years to meet her renovation goal.