# If $600 is deposited in an account paying 8.5% annual interest, compounded continuously, how long will it take for the account to increase to$800?

Oct 13, 2015

$\log \frac{\frac{4}{3}}{\log} \left(1.085\right) \approx 3.52638$ years $\approx 1288$ days

#### Explanation:

This answer reflects my understanding of compound interest. It may be that the correct rate for continuously compounded interest results in a higher effective annual interest rate.

Since adding 8.5% really means multiplying by $\frac{100 + 8.5}{100} = 1.085$, we want to solve:

$600 \cdot {1.085}^{t} = 800$

Divide both sides by $600$ to get:

${1.085}^{t} = \frac{4}{3}$

Take logs of both sides to get:

$t \log \left(1.085\right) = \log \left(\frac{4}{3}\right)$

Divide both sides by $\log \left(1.085\right)$ to get:

$t = \log \frac{\frac{4}{3}}{\log} \left(1.085\right) \approx 3.52638$ years $\approx 1288$ days

Oct 13, 2015

$t = 3.35 Y e a r s$

#### Explanation:

This can be solved using continuous compound interest formula

$A = p {e}^{r t}$

p = principle interest
r = annual interest
t = number of years
A = Amount after n years including interest

Here,

p = 600
r = 8.5 / 100 = 0.085
t = 800

so,

$800 = 600. {e}^{0.085 . t}$
$\frac{800}{6000} = {e}^{0.085 t}$
$1.33 = {e}^{0.085 t}$

Taking natural log(ln) on both sides
$\ln \left(1.33\right) = \ln \left({e}^{0.085 t}\right)$
$0.285 = 0.085 t$
$t = 3.35 Y e a r s$