You invest $25 in an account earning 7.5% interest per year. After how many years is it worth $42.50?
The general formula for compound interest is:
F=P*(1+i)^n
i=interest per annum, 0.075
n=number of years
F=42.50
P=25
42.5=25(1.075)^n
(1.075)^n=42.5/25=1.7
take log on both sides,
nlog(1.075)=log(1.7)
n=log(1.7)/log(1.075)=7.337 years
That can't be right...the correct answer shows it should be 12 years.
Please check your figures and your answers.
My answer, figured the long way, is similar to MathMate's.
I have already applied a quick check using the rule of 72, which states that money doubles every 72/i% years.
In this case, the number of years to double is 72/7.5=9.6 years approximately. So 12 years cannot be right unless there is a typo somewhere.
Are you sure the interest rate is not 4.5% compounded annually?
To find out how many years it takes for an investment to grow from $25 to $42.50 with an interest rate of 7.5%, we can use the formula for compound interest:
A = P * (1 + r/n)^(n*t)
Where:
A = the future value of the investment
P = the principal amount (initial investment)
r = the annual interest rate (as a decimal)
n = the number of times the interest is compounded per year
t = the number of years
In this case, we want to find t, so we rearrange the formula:
t = log(A/P) / log(1 + r/n)
Plugging in the given values:
A = $42.50
P = $25
r = 0.075 (7.5% as a decimal)
n = 1 (compounded annually)
t = log(42.50/25) / log(1 + 0.075/1)
To simplify the calculation, we can use a calculator or a software program that supports logarithmic functions. Evaluating the expression:
t ≈ log(1.7) / log(1.075)
t ≈ 0.2304489214 / 0.0314204544
t ≈ 7.324180211
After solving the equation, we find that it takes approximately 7.32 years for the investment to grow from $25 to $42.50 with an interest rate of 7.5%.