I need help with this problem

[waunar]

An investment of P dollars increased to A dollars in t years. If interest was compounded contiouously, find the interest rate.

A=13,464

P=1000

t=20

r= unkown
I can solve the problem if the they ask me for A p or t but i don't know how to solve for r.


another problem i am having difficulties with is as follow.
Graph F on the interval (0, 200). find an approimate equation for the horizontal asymptote.

f(x)=(1+(1/x))x the x is an exponent

 

[tkhunny]

An investment of P dollars increased to A dollars in t years. If interest was compounded contiouously, find the interest rate.
A=13,464
P=1000
t=20
r= unkown
I can solve the problem if the they ask me for A p or t but i don't know how to solve for r.

You'll need logarithms. Have you useen those?

Graph F on the interval (0, 200). find an approximate equation for the horizontal asymptote.

f(x)=(1+(1/x))x the x is an exponent

What is your plan? Did you graph it? Can we use calculus?

 

[soroban]

Hello, waunar!

An investment of P dollars increased to A dollars in t years.
If interest was compounded contiouously, find the interest rate.
A = 13,464, P = 1000, t = 20, r = unknown

I can solve the problem if the they ask me for A p or t but i don't know how to solve for r.

Hmmm, you say you know hot to solve for $\displaystyle t$, right? . . . I don't see any problem here.

We have: \(\displaystyle \:A\:=\e^{rt}\;\;\Rightarrow\;\;13,464\:=\:1000e^{20r}\)

Use the same method you used to solve for $\displaystyle t$ . . .

Divide by 1000: $\displaystyle \;e^{20r}\:=\:13.464$

Take logs: $\displaystyle \;\ln(e^{20r})\:=\:\ln(13.464)\;\;\Rightarrow\;\;20r\cdot\ln(e)\:=\:\ln(13.464)\;\;\Rightarrow\;\;20r\:=\:\ln(13.464)$

Hence: $\displaystyle \:r\:=\:\frac{\ln(13.464)}{20}\:=\:0.130000973$

$\displaystyle \;\;$Therefore: $\displaystyle \,r\:=\:13\%$

Graph $\displaystyle f$ on the interval (0, 200).
Find an approximate equation for the horizontal asymptote.

$\displaystyle \;\;\;f(x)\;=\;\left(1\,+\,\frac{1}{x}\right)^x$

Are we expected to look at the graph and "eyeball" the asymptote?
$\displaystyle \;\;$Kind of primitive, ain't it?


A horizontal asymptote occurs if $\displaystyle \,\lim_{x\to\infty}f(x)$ has a finite limit.

It happens that: $\displaystyle \,\lim_{x\to\infty}\left(1\,+\,\frac{1}{x}\right)^x\,$ is the definition of $\displaystyle e$.

Therefore, the horizontal asymptote is: $\displaystyle \,y\:=\:e$