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- Thread starter Juliayaho
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\(\displaystyle \frac{pr}{x}=1-(1+r)^n\)

I would arrange this as:

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

Can you convert this from exponential to logarithmic form or take the natural or base 10 logarithm of both sides?

- Aug 30, 2012

- 1,123

You've got a good start.help me out... Im 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!!

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

[tex]pr = x \left ( 1 - (1 + r)^n \right )[/tex]

[tex]\frac{pr}{x} = 1 - (1 + r)^n[/tex]

Now isolate the term with the n in it.

[tex]\frac{pr}{x} - 1 = - (1 + r)^n[/tex]

[tex]1 -\frac{pr}{x} = (1 + r)^n[/tex]

What comes next?

-Dan

Edit: Agh. MarkFL beat me to it. But my work is prettier....(flower)

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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 part

Thank you all by the way!

You've got a good start.

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

[tex]pr = x \left ( 1 - (1 + r)^n \right )[/tex]

[tex]\frac{pr}{x} = 1 - (1 + r)^n[/tex]

Now isolate the term with the n in it.

[tex]\frac{pr}{x} - 1 = - (1 + r)^n[/tex]

[tex]1 -\frac{pr}{x} = (1 + r)^n[/tex]

What comes next?

-Dan

Edit: Agh. MarkFL beat me to it. But my work is prettier....(flower)

- Jun 30, 2012

- 176

$\text{My humble offering: }$

[FONT=MathJax_Main]$1-\frac{pr}{x}=(1+r)^{n}$[/FONT]

$\text{Having reached this point, and following the general principle that, to solve}$

$\text{exponential equations, one usually}$

$\text{needs to take logs of both sides, the next move is: }$

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

$$\text{To those patiently helping this member, please forgive me if I}$$

$$\text{seem to be coming in over}$$

$$\text{the top: this is the first time I've felt vaguely}$$

$$\text{confident in being able to help someone since I}$$

$$\text{joined. My enthusiasm has overmastered my reticence.}$$

[FONT=MathJax_Main]$1-\frac{pr}{x}=(1+r)^{n}$[/FONT]

$\text{Having reached this point, and following the general principle that, to solve}$

$\text{exponential equations, one usually}$

$\text{needs to take logs of both sides, the next move is: }$

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

$$\text{To those patiently helping this member, please forgive me if I}$$

$$\text{seem to be coming in over}$$

$$\text{the top: this is the first time I've felt vaguely}$$

$$\text{confident in being able to help someone since I}$$

$$\text{joined. My enthusiasm has overmastered my reticence.}$$

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