MHB Lost in Exponential Equations? Need Help Solving for n?

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The discussion focuses on solving the equation p = x*(1-(1+r)^n)/r for n. Participants guide the original poster through the algebraic manipulation needed to isolate the term involving n, ultimately leading to the equation 1 - (pr/x) = (1 + r)^n. They suggest using logarithms to solve for n, specifically n = ln(1 - pr/x) / ln(1 + r). The original poster is working on a loan calculator project and needs this solution to determine the number of months required to pay off a loan. The community offers support and encouragement throughout the problem-solving process.
Juliayaho
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help me out... I am trying to solve for n
p=x*(1-(1+r)^n)/r

so far I know p*r=x*(1-(1+r)^-n)
then (p*r)/x=(1-(1+r)^n)

but then i get lost... help!

thanks!
 
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You have gotten as far as:

$$\frac{pr}{x}=1-(1+r)^n$$

I would arrange this as:

$$(1+r)^n=1-\frac{pr}{x}$$

Can you convert this from exponential to logarithmic form or take the natural or base 10 logarithm of both sides?
 
Juliayaho said:
help me out... I am trying to solve for n
p=x*(1-(1+r)^n)/r

so far I know p*r=x*(1-(1+r)^n)
then (p*r)/x=(1-(1+r)^n)

but then i get lost... help!

thanks!
You've got a good start.

p = \frac{x \left ( 1 - (1 + r)^n \right )}{r}

pr = x \left ( 1 - (1 + r)^n \right )

\frac{pr}{x} = 1 - (1 + r)^n

Now isolate the term with the n in it.
\frac{pr}{x} - 1 = - (1 + r)^n

1 -\frac{pr}{x} = (1 + r)^n

What comes next?

-Dan

Edit: Agh. MarkFL beat me to it. But my work is prettier...(flower)
 
Yeah that's exactly what I don't remember the "(1 + r)^-n" part :/
I need to solve for n because it represents the number of month it would take me to payoff a loan... Is for a project that I'm doing... It's a one mor detail I would add to my loan calculator.. So I need to use the formula as n=
But I don't remember that logarithm partThank you all by the way!
topsquark said:
You've got a good start.

p = \frac{x \left ( 1 - (1 + r)^n \right )}{r}

pr = x \left ( 1 - (1 + r)^n \right )

\frac{pr}{x} = 1 - (1 + r)^n

Now isolate the term with the n in it.
\frac{pr}{x} - 1 = - (1 + r)^n

1 -\frac{pr}{x} = (1 + r)^n

What comes next?

-Dan

Edit: Agh. MarkFL beat me to it. But my work is prettier...(flower)
 
topsquark said:
Edit: Agh. MarkFL beat me to it. But my work is prettier...(flower)
Seems like it's a contest about who is first :P This time I win :)

try read this, if you solve it or still struggle we will help you:)!

Regards
$$|\pi\rangle$$
 
$\text{My humble offering: }$
[FONT=MathJax_Main]$1-\frac{pr}{x}=(1+r)^{n}$
$\text{Having reached this point, and following the general principle that, to solve}$
$\text{exponential equations, one usually}$
$\text{needs to take logs of both sides, the next move is: }$

$$n\cdot ln(1+r)=ln(1-\frac{pr}{x})$$

$$\text{To those patiently helping this member, please forgive me if I}$$
$$\text{seem to be coming in over}$$
$$\text{the top: this is the first time I've felt vaguely}$$
$$\text{confident in being able to help someone since I}$$
$$\text{joined. My enthusiasm has overmastered my reticence.}$$
 
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