Lost in Exponential Equations? Need Help Solving for n?

[Juliayaho]

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!

 

[MarkFL]

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?

 

[topsquark]

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)

 

[Juliayaho]

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!

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)

 

[Petrus]

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
$$|\pi\rangle$$

 

[DeusAbscondus]

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

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

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