Question

A deposit earns interest at a rate of r percent compounded continuously and doubles in value in 9 years. Find r. (Round your answer to two decimal places.) g

Answer 1

Answer:

The earning rate is approximately 0.08.

Step-by-step explanation:

We can determine the yearly rate by means of compound interest, which is defined by:

$C(t) = C_{o}\cdot (1+r)^{t}$ (Eq. 1)

Where:

$C_{o}$ - Initial deposit, measured in US dollars.

$r$ - Earning rate, dimensionless.

$t$ - Earning periods, measured in years.

We proceed to clear the earning rate within:

$\frac{C(t)}{C_{o}} = (1+r)^{t}$

$\log \frac{C(t)}{C_{o}} = t\cdot \log (1+r)$

$\frac{1}{t}\cdot \log \frac{C(t)}{C_{o}} = \log (1+r)$

$\log \left(\frac{C(t)}{C_{o}} \right)^{\frac{1}{t} } = \log (1+r)$

$\left(\frac{C(t)}{C_{o}} \right)^{\frac{1}{t} } = 1+r$

$r = \left(\frac{C(t)}{C_{o}} \right)^{\frac{1}{t} }-1$

If we know that $C(9) = 2\cdot C_{o}$ and $t = 9$, then the earning rate is:

$r = 2^{\frac{1}{9} }-1$

$r \approx 0.08$

The earning rate is approximately 0.08.