# Math Help - Bank Compounding

1. ## Bank Compounding

I can't seem to get these problems solved. I have attempted to follow examples given, but I continue to get stuck at a certain point in the process. Please can someone help? Thanks kindly.

Bank Account

If a bank compounds continuous, then the formula becomes simpler, that is A=P e^(rt) where e is a constant and equals approximately 2.7183. Calculate A with continuous compounding. Round your answer to the hundredth's place.

Now suppose, instead of knowing t, we know that the bank returned to us $25,000 with the bank compounding continuously. Using logarithms, find how long we left the money in the bank (find t). Round your answer to the hundredth's place. A commonly asked question is, "How long will it take to double my money?" At 8% interest rate and continuous compounding, what is the answer? Round your answer to the hundredth's place. 2. Originally Posted by fw_mathis I can't seem to get these problems solved. I have attempted to follow examples given, but I continue to get stuck at a certain point in the process. Please can someone help? Thanks kindly. Bank Account If a bank compounds continuous, then the formula becomes simpler, that is A=P e^(rt) where e is a constant and equals approximately 2.7183. Calculate A with continuous compounding. Round your answer to the hundredth's place. Now suppose, instead of knowing t, we know that the bank returned to us$25,000 with the bank compounding continuously. Using logarithms, find how long we left the money in the bank (find t). Round your answer to the hundredth's place.

A commonly asked question is, "How long will it take to double my money?" At 8% interest rate and continuous compounding, what is the answer? Round your answer to the hundredth's place.
By the look of this you have missed some data out.

RonL

3. Hello, fw_mathis!

You left out most of the problem . . .

If a bank compounds continuously, the formula becomes simpler.
That is: . $A \:=\:Pe^{rt}$ where $e$ is a constant,
. . $r$ is the annual interest rate and $t$ is the number of years.

Calculate $A$ with continuous compounding.

I assume this is the last half of your previous post.
Then: . $P = 20,\!000,\;r = 0.08,\;t = 3$ . . . right?

We have: . $A \;=\;20,\!000e^{(0.08)(3)}\;\approx\;\boxed{\25,\ !424.98}$

Now suppose, instead of knowing $t$, we know that the bank
Using logarithms, find how long we left the money in the bank (find $t$).

The equation becomes: . $25,\!000 \:=\:20,\!000e^{0.8t}\quad\Rightarrow\quad e^{0.08t}\:=\:1.25$

Take logs: . $\ln\left(e^{0.08t}\right) \:=\:\ln(1.25)\quad\Rightarrow\quad 0.08t\ln(e) \:=\:\ln(1.25)$

Therefore: . $0.08t\:=\:\ln(1.25)\quad\Rightarrow\quad t \:=\:\frac{\ln(1.25)}{0.08}\:\approx\:\boxed{2.79\ text{ years}}$

A commonly asked question is, "How long will it take to double my money?"
At 8% interest rate and continuous compounding, what is the answer?
We want our $\20,\!000$ to grow to $\40,\!000.$
The equation is: . $40,\!000 \:=\:20,\!000e^{0.08t}\quad\Rightarrow\quad e^{0.08t} \:=\:2$
Take logs: . $\ln\left(e^{0.08t}\right) \:=\:\ln 2\quad\Rightarrow\quad 0.08t\ln(e)\:=\:\ln 2$
Therefore: . $0.08t\:=\:\ln 2\quad\Rightarrow\quad t\:=\:\frac{\ln 2}{0.08}\:\approx\:8.66\text{ years}$