1. ## Continuous compounding

I'm having a lot of trouble with my calculus homework. Does anyone know how to work this problem?
Suppose the consumption of electricity grows at 7.6% per year, compounded continuously. Find the number of years before the use of electricity has tripled. Round the answer to the nearest hundredth.

14.46 , 39.47 , 0.14 , 1.45

2. ANYBODY??? I figured the answer but am having problems with the formula. I came up with fv=p(1+r/n)^t, and cannot figure out t....my formula looks like this:

3=1(1.076)^t

help...........

3. I got 14.55 years - I will post how I did it in a minute - I have to get the equations into this box

a(t) = lim as n approaches infinity ( a0(1+ r/n)^(nt) )
where a0 is the initial amount of power used, r is the rate of growth (.076) and n is the number of times it is compounded

Then you use L'Hopital's rule to find that the above function is the same as this

a(t) = a0e^(rt)

(Hopefully this helps a little)

$
\mathop {\lim }\limits_{n \to \infty } \ln ((1 + \frac{r}
{n})^{nt} ) = \mathop {\lim }\limits_{n \to \infty } (nt)\ln (1 + \frac{r}
{n}) = \mathop {\lim }\limits_{n \to \infty } t \bullet \frac{{\ln (1 + \frac{r}
{n})}}
{{1/n}}
$

$
\begin{gathered}
= \mathop {\lim }\limits_{n \to \infty } t \bullet \frac{{\frac{1}
{{1 + \frac{r}
{n}}}}}
{{ - 1/n^2 }}( - r/n^2 ) = \mathop {\lim }\limits_{n \to \infty } t\frac{n}
{{n + r}}(r) = \frac{1}
{1}rt \hfill \\
\mathop {\lim }\limits_{n \to \infty } e^{\ln f(x)} = e^L = e^{rt} \hfill \\
\end{gathered}
$

a(t) = 3 = 1*e^(.076*t)
ln(3) = .076*t
t=ln(3)/.076 = 14.55 years

4. Anytime something is compounded continuously, use the $A=Ce^{kt}$ formula. Or, if it is money that is being compounded continuously, use $A=Pe^{rt}$ formula. Either way, the formulas are equivalent.

Or like billa said, you can use the regular compound formula and just find the ${\lim x \rightarrow \infty}$.