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

  • Thread starter Thread starter Juliayaho
  • Start date Start date
Juliayaho
Messages
13
Reaction score
0
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!
 
Last edited:
Mathematics news on Phys.org
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]$1-\frac{pr}{x}=(1+r)^{n}$
$\text{Having reached this point, and following the general principle that, to solve}$
$\text{exponential equations, one usually}$
$\text{needs to take logs of both sides, the next move is: }$

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

$$\text{To those patiently helping this member, please forgive me if I}$$
$$\text{seem to be coming in over}$$
$$\text{the top: this is the first time I've felt vaguely}$$
$$\text{confident in being able to help someone since I}$$
$$\text{joined. My enthusiasm has overmastered my reticence.}$$
 
Last edited:
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top