Need serious help with midtern review questions.

Yelfett2 · #1 $Old$ November 5th, 2009, 07:51 PM

serious issues with some problems.

1. The ABC corporation issues a bond that pays 10,000$ after 10 years. If the initial investment is 5,000$ what is the interest rate per year, compounded annually.

The equation to use is A = P ( 1 + r )^t

Where A = Ending amount
P = Starting amount
r = interest rate
t = time

Right now I have this
10,000 = 5,000 ( 1 + r ) ^ 10
2 = ( 1 + r ) ^ 10
$Log_((1+r)2=10$

But i'm not sure where to go from here. Any ideas?

2. Mary wants to invest money in a 7-year saving certificate with an interest rate of 6% per year, compounded continuously. How much will she need to invest to have $1000 at the end of the term.

This uses the equation A = Pe^rt Where all variables have the same value is before, and of course e is e (2.7.........)

So far I have
1000=Pe^(7 x .06)
1000=Pe^.42
$ln(1000)/ln(.42)=lnP$

And now I have no idea what to do

3. Write the estimate in exponential form (Using the definition of a logarithm)

$ln10 = 2.303$

With this one I have absolutely no idea what I should do.

4. A water filter is constructed from a foam that REMOVES 25% of impurities for each inch of foam. How many inches of foam are needed for the filter to remove 50% of the impurities.

I know that this can be modeled with $A = Pa^x$ and also with A = Pe^(rx)

P = concentration of impurities before filtration
A = concentration of impurities in filtered water
x = thickness of foam

I understand that $.5P = Pa^x$

But I can't move much further from there.

Please don't just give answers. I really need to understand how to solve each problem. Thank you in advance.

mathaddict · #2 $Old$ November 5th, 2009, 08:33 PM

Quote:

Originally Posted by Yelfett2 $View Post$

serious issues with some problems.

1. The ABC corporation issues a bond that pays 10,000$ after 10 years. If the initial investment is 5,000$ what is the interest rate per year, compounded annually.

The equation to use is A = P ( 1 + r )^t

Where A = Ending amount
P = Starting amount
r = interest rate
t = time

Right now I have this
10,000 = 5,000 ( 1 + r ) ^ 10
2 = ( 1 + r ) ^ 10
$Log_((1+r)2=10$

But i'm not sure where to go from here. Any ideas?

2. Mary wants to invest money in a 7-year saving certificate with an interest rate of 6% per year, compounded continuously. How much will she need to invest to have $1000 at the end of the term.

This uses the equation A = Pe^rt Where all variables have the same value is before, and of course e is e (2.7.........)

So far I have
1000=Pe^(7 x .06)
1000=Pe^.42
$ln(1000)/ln(.42)=lnP$

And now I have no idea what to do

3. Write the estimate in exponential form (Using the definition of a logarithm)

$ln10 = 2.303$

With this one I have absolutely no idea what I should do.

4. A water filter is constructed from a foam that REMOVES 25% of impurities for each inch of foam. How many inches of foam are needed for the filter to remove 50% of the impurities.

I know that this can be modeled with $A = Pa^x$ and also with A = Pe^(rx)

P = concentration of impurities before filtration
A = concentration of impurities in filtered water
x = thickness of foam

I understand that $.5P = Pa^x$

But I can't move much further from there.

Please don't just give answers. I really need to understand how to solve each problem. Thank you in advance.

HI

(1) You can rewrite 2 as $(2^{\frac{1}{10}})^10$

$(2^{\frac{1}{10}})^10=(1+r)^10$

Comparing , $2^{\frac{1}{10}}=1+r$

$r=2^{\frac{1}{10}}-1$

(2) $1000 =Pe^{0.42}$

$P=\frac{1000}{e^0.42}$

where here you could use your calculator since e has a value instead of taking the logs .

(3) $a^x=y\Rightarrow log_a{y}=x$

so here $\ln 10=2.303$ ..... $e^{2.303}=10$

By the way , you should only post one question per thread . This is one of the rules of the forum.

Yelfett2 · #3 $Old$ November 5th, 2009, 09:15 PM

Quote:

Originally Posted by mathaddict $View Post$

HI

(1) You can rewrite 2 as $(2^{\frac{1}{10}})^10$

$(2^{\frac{1}{10}})^10=(1+r)^10$

Comparing , $2^{\frac{1}{10}}=1+r$

$r=2^{\frac{1}{10}}-1$

(2) $1000 =Pe^{0.42}$

$P=\frac{1000}{e^0.42}$

where here you could use your calculator since e has a value instead of taking the logs .

(3) $a^x=y\Rightarrow log_a{y}=x$

so here $\ln 10=2.303$ ..... $e^{2.303}=10$

By the way , you should only post one question per thread . This is one of the rules of the forum.

Thank you.

Thats a rather odd rule dont you think? Better to have 1 thread with 4 questions than 4 threads with 1 question.

Wilmer · #4 $Old$ November 5th, 2009, 09:50 PM

Quote:

Originally Posted by Yelfett2 $View Post$

Thats a rather odd rule dont you think? Better to have 1 thread with 4 questions than 4 threads with 1 question.

Well, when you start your own site, then put in that rule.
To each his own.

mr fantastic · #5 $Old$ November 6th, 2009, 03:46 PM

Quote:

Originally Posted by Yelfett2 $View Post$

Thank you.

Thats a rather odd rule dont you think? Better to have 1 thread with 4 questions than 4 threads with 1 question.

Think about it. Threads can get convoluted and difficult to follow (from all the replies) when there are too many questions.