Your final balance of an investment of $270 invested at 10% was $459. For what period of time did you invest?
Thanks but why did you add 260 twice
To determine the period of time you invested, we can use the formula for compound interest:
A = P (1 + r/n)^(nt)
Where:
A = final balance
P = principal amount (initial investment)
r = annual interest rate
n = number of times interest is compounded per year
t = number of years
In this case, we know:
A = $459
P = $270
r = 10% = 0.10
To find the value of t, we need to rearrange the formula:
A/P = (1 + r/n)^(nt)
Substituting the given values:
459/270 = (1 + 0.10/n)^(nt)
Now we can solve for t. It involves some algebraic manipulation, so here are the steps:
1. Divide both sides by 270:
459/270 = (1 + 0.10/n)^(nt)
2. Take the natural logarithm (ln) of both sides to remove the exponent:
ln(459/270) = ln[(1 + 0.10/n)^(nt)]
3. Apply the power rule of logarithms:
ln(459/270) = (nt) * ln(1 + 0.10/n)
4. Divide both sides by ln(1 + 0.10/n):
ln(459/270) / ln(1 + 0.10/n) = nt
5. Divide both sides by n:
ln(459/270) / (n * ln(1 + 0.10/n)) = t
Now we have an expression for t.
To find the exact value, we need to know the compounding frequency (n). If it's not specified, we'll assume annual compounding (n = 1).
Let's calculate t using a calculator:
ln(459/270) / (1 * ln(1 + 0.10/1)) ≈ 4.24 years
Therefore, you invested for approximately 4.24 years.
P = Po + Po*r*t.
459 = 270 + 270*0.1*t
t = ?. = Time in years.
In factored form:
P = Po + Po*r*t = Po(1+rt).