Solve for n when n is an exponent

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zingbust

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I can't figure out how to do this...
Solve for n: A=P(i(1+i)^n)/(((1+i)^n)-1)
Note: i has nothing to do with imaginary number, it means interest rate.
Anyone? Thanks.mortgageProblem.png
 
zingbust said:
I can't figure out how to do this...
Solve for n: A=P(i(1+i)^n)/(((1+i)^n)-1)
Note: i has nothing to do with imaginary number, it means interest rate.
Anyone? Thanks.View attachment 33288
Click to expand...
As @topsquark said, solve for (1+i)^n - this should be quite simple

You should get:
[math](1+i)^n=\frac{A}{A-iP}[/math]
At this point take logs of both sides... and recall [math]log(x^n)=nlog(x)[/math]
Your answer is:
[math]n=\frac{\log(\frac{A}{A-iP})}{\log(1+i)}[/math]
 
BigBeachBanana said:
Change of base formula:
\(\displaystyle n=\log_{1+i}\left(\frac{A}{A-iP}\right)\)
Click to expand...

ausmathgenius420 said:
As @topsquark said, solve for (1+i)^n - this should be quite simple

You should get:
[math](1+i)^n=\frac{A}{A-iP}[/math]
At this point take logs of both sides... and recall [math]log(x^n)=nlog(x)[/math]
Your answer is:
[math]n=\frac{\log(\frac{A}{A-iP})}{\log(1+i)}[/math]
Click to expand...
Yes, thanks that is quite simple: n = sq rt of (1 + i)
Thank you. Now I'm plugging in A, i and P for my particular case to see if it makes sense.
 
Unless I did something wrong, I am getting n = .002778 which doesn't make sense since in my case A = Amount of monthly payment (Principal and Interest only) which is $877.19 and i = interest rate (monthly) expressed as a decimal. Annual interest rate is 5.375% which is .05375 which is divided by 12 to get monthly = 0.0044791666666667 and P = Principal loan balance = $59,028.01.

OK, so iP = 264.3962947916686 and A - iP = 612.7937052083314 and A / (A - iP) = 1.431460526673951 and finally log ( 1 + i ) = 0.00194093372 so .00194093372 times 1.431460526673951 = 0.0027783700050704 which is n.
However n is the number of remaining monthly payments to fully amortize the loan therefore it has to be somewhere in the range of 6 1/2 to 8 years which is between 78 and 96. And in conclusion, n should be somewhere in the range of 78 to 96.

Sorry I showed all the very simple math, but I wanted to get a math expert who is also fully aware of amortization of loans to confirm what I said before I move on to the tough stuff. Don't show me the answer or you'll be reported by lookagain!! Just tell me if I'm on the right track!
Thanks.
 
zingbust said:
Unless I did something wrong, I am getting n = .002778 which doesn't make sense since in my case A = Amount of monthly payment (Principal and Interest only) which is $877.19 and i = interest rate (monthly) expressed as a decimal. Annual interest rate is 5.375% which is .05375 which is divided by 12 to get monthly = 0.0044791666666667 and P = Principal loan balance = $59,028.01.

OK, so iP = 264.3962947916686 and A - iP = 612.7937052083314 and A / (A - iP) = 1.431460526673951 and finally log ( 1 + i ) = 0.00194093372 so .00194093372 times 1.431460526673951 = 0.0027783700050704 which is n.
However n is the number of remaining monthly payments to fully amortize the loan therefore it has to be somewhere in the range of 6 1/2 to 8 years which is between 78 and 96. And in conclusion, n should be somewhere in the range of 78 to 96.

Sorry I showed all the very simple math, but I wanted to get a math expert who is also fully aware of amortization of loans to confirm what I said before I move on to the tough stuff. Don't show me the answer or you'll be reported by lookagain!! Just tell me if I'm on the right track!
Thanks.
Click to expand...
You didn't use the formula correctly.
\(\displaystyle n=\log_{1+i}\left(\frac{A}{A-iP}\right)\)
This is log with base of (1+i), similar to log with base e is the natural log, ln. It is not multiplication as you did.
If you're still confused. Use this formula instead, it's the equivalent.
\(\displaystyle n=\frac{\log(\frac{A}{A-iP})}{\log(1+i)}\)

From your numbers, I got [imath]n \approx 80.26[/imath].
 
Here is my work so far. I had a friend help.
For simplicity I am going to let x = (1+i)^n and get rid of the x later by putting the (1+i)^n back in
Start by expressing the original equation with P in the numerator of the right side instead of leaving it separate....
A = Pix/x -1
Multiply both sides by the denominator
A(x - 1) = Pix
and now divide both sides by Pi
A(x-1)/Pi = x
now reverse the 2 sides
x = A(x-1)/Pi
now take the log of both sides
log(x) = log[A(x-1)]/Pi
now and only now substitute the x for (1+i)^n
log[(1+i)^n] = log[A[(1+i)^n]/Pi

and this is where we stopped because n is still on both sides of the equation.
Trying to solve for just (1+i)^n, the best I could do was try....
n = log (1+i), is that right?

Thanks for responses.
 

Attachments

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zingbust said:
Here is my work so far. I had a friend help.
For simplicity I am going to let x = (1+i)^n and get rid of the x later by putting the (1+i)^n back in
Start by expressing the original equation with P in the numerator of the right side instead of leaving it separate....
A = Pix/x -1
Multiply both sides by the denominator
A(x - 1) = Pix
and now divide both sides by Pi
A(x-1)/Pi = x
now reverse the 2 sides
x = A(x-1)/Pi
now take the log of both sides
log(x) = log[A(x-1)]/Pi
now and only now substitute the x for (1+i)^n
log[(1+i)^n] = log[A[(1+i)^n]/Pi

and this is where we stopped because n is still on both sides of the equation.
Trying to solve for just (1+i)^n, the best I could do was try....
n = log (1+i), is that right?

Thanks for responses.
Click to expand...
The problem occurs at x = A(x-1)/Pi, where you have x on both sides, so you are not ready to take the log.

First finish solving for x: expand the right side, and collect terms containing x on the left. (Actually, I'd do that back at A(x - 1) = Pix, to avoid fractions.) Then factor so you have something times x, and divide by that something. Then you will have x alone, and can replace it with (1+i)^n and solve for n (using the log).
 
OK, but first things first. I was asked to start by solving (1 + i)^n and I now believe the answer is n log(1 + i) which is not really "solving" anything, just expressing the expression in a different form so that it will make it easier to work on the main problem of solving for n in the original formula.
 
zingbust said:
OK, but first things first. I was asked to start by solving (1 + i)^n and I now believe the answer is n log(1 + i) which is not really "solving" anything, just expressing the expression in a different form so that it will make it easier to work on the main problem of solving for n in the original formula.
Click to expand...
Follow the instructions provided in response #11 - with pencil/paper - write your calculations mathematically.
 
zingbust said:
OK, but first things first. I was asked to start by solving (1 + i)^n and I now believe the answer is n log(1 + i) which is not really "solving" anything, just expressing the expression in a different form so that it will make it easier to work on the main problem of solving for n in the original formula.
Click to expand...
That's not the first thing; that's the last thing! Do what I suggested, and then what you say fits in here:
Dr.Peterson said:
Then you will have x alone, and can replace it with (1+i)^n and solve for n (using the log).
Click to expand...
Before you do that, you need to have an equation that says "x = ...". Then when you replace x with what it means, you'll have "(1 + i)^n = ...", and can then take the log of both sides.

You are evidently misunderstanding this:
topsquark said:
Start by solving for (1 + i)^n...

-Dan
Click to expand...
What he means is to write the equation as "(1 + i)^n = ...", isolating that expression, which is identical to solving for your x (which is a very good idea that you suggested, in #10). "Solving for" something is not the same as "solving" something.
 
OK, I understand everything everyone said except for how to arrive at the first solution in #9. I arrived at the 2nd solution in #9 like this....
First the algebra using x = (1+i)^n... starting with ...
A(x - 1) = Pix ... as suggested by Dr. Peterson in #11
(Ax - A)/x = Pi ... after re-expressing the left side and dividing both sides by x

now re-express left side again to get a common denominator and switch P and i around on the right
Ax/x - A/x = iP

the 2 x's in the left part of the left side cancel so now we have
A - A/x = iP

now subtract both sides by A
-A/x = iP - A

and now take reciprocals of both sides
x/-A = 1/(iP -A)

multiply both sides by -A
x = -A/(iP -A) .... and we have solved for x but let's make it simpler and easier to read

multiply top and bottom of right side by -1 which is the same as multiplying all of right side by 1
x = A/-1(iP - A)

expand bottom of right side
x = A/-iP + A

which is the same as
x = A/A- iP

and now and only now substitute x for (i + 1)^n
(i + 1)^n = A/A -iP ... so that we're ready to take the logs (see attached for my work on the logs part)

Comments appreciated especially if I arrived at the correct formula completely by accident.
 

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