Alright, this is a bonus question on my homework. I think I did it wrong...
How long will it take for an investment of $1000 to grow to $1750 if interest is compounded continuously at a rate of 8%? Round to the nearest tenth.
A=Pe^rt
$1750=$1083.38
divided
ans= 1.62
yes, you did it wrong. You left out the t part:
1000 e^.08t = 1750
e^.08t = 1.75
.08t = log(1.75)
t = log(1.75)/.08 = 6.99
or, 7 years.
surely you didn't think that the money wold almost double in a year and a half at 8%!
I did indeed. Thank you again for helping me out! Have a great night!
To calculate the time it takes for an investment to grow to a certain value with continuous compounding interest, you can use the formula A = Pe^rt. In this formula, A represents the final amount, P represents the initial principal (investment amount), e is the mathematical constant approximately equal to 2.71828, r represents the interest rate, and t represents the time in years.
In your case, the initial principal (P) is $1000, the final amount (A) is $1750, and the interest rate (r) is 8% or 0.08 (expressed as a decimal).
The formula becomes:
$1750 = $1000 * e^(0.08t)
To isolate the variable t, you need to divide both sides of the equation by $1000:
e^(0.08t) = $1750 / $1000
Simplifying further:
e^(0.08t) = 1.75
To solve for t, you need to take the natural logarithm (ln) of both sides of the equation:
ln(e^(0.08t)) = ln(1.75)
Using the logarithmic property, the exponent (0.08t) can be brought down as a coefficient:
0.08t * ln(e) = ln(1.75)
Since ln(e) equals 1, the equation simplifies to:
0.08t = ln(1.75)
Finally, divide both sides of the equation by 0.08 to solve for t:
t = ln(1.75) / 0.08
Using a calculator to compute this, you would find:
t ≈ 4.41 years (rounded to two decimal places)
So, it would take approximately 4.41 years for the investment of $1000 to grow to $1750 with continuous compounding at an interest rate of 8%.