An equation with unknown exponent

**p.numminen** · January 28th 2008, 07:49 AM

Is is possible to solve this equation?

$0.72 \times 1.04^{n} + 0.28 = 1.035568^{n}$

If so, how?

**Peritus** · January 28th 2008, 11:23 AM

several ways come to mind:

1. Newton's method - Wikipedia, the free encyclopedia

2. Bisection method - Wikipedia, the free encyclopedia

3. Secant method - Wikipedia, the free encyclopedia

4. you can also plot the two functions on each side of the equation on the same system of axis, and find their intersection point.

5. you can try to converge to the solution:

$ 0.72 \times 1.04^{n} + 0.28 = 1.035568^{n} $

$ \Leftrightarrow n_{k + 1} = \frac{{\ln \left( {0.72 \times 1.04^{n_k } + 0.28} \right)}} {{\ln 1.035568}} $

the stopping condition would be: $ \left| {n_{k + 1} - n_k } \right| < \varepsilon $

in this case one obvious solution is n = 0

**earboth** · January 28th 2008, 09:12 PM

Originally Posted by p.numminen

Is is possible to solve this equation?

$0.72 \times 1.04^{n} + 0.28 = 1.035568^{n}$

If so, how?

It would be interesting for us to know where you got your values from.

If the 1.04 is the rounded value of 1.035568 then we can label this constant as c and your equation becomes:

$0.72 \cdot c^n + 0.28 = c^n~\iff~0.28 = 0.28 \cdot c^n ~\iff~ 1 = c^n~\implies~ n = 0$

**p.numminen** · January 29th 2008, 05:45 AM

No, 1.04 is not the rounded value of 1.035568. They are both exact numbers, actually.

I made the equation for a practical problem, but then I realised that I cannot solve the equation myself.

**colby2152** · January 29th 2008, 06:39 AM

Originally Posted by p.numminen

Is is possible to solve this equation?

$0.72 \times 1.04^{n} + 0.28 = 1.035568^{n}$

If so, how?

As the others have mentioned, an obvious solution is $n=0$

$0.72 \times 1.04^{0} + 0.28 = 1.035568^{0}$

$0.72 + 0.28 = 1$

dennis · February 20th 2008, 06:29 AM

I guess A(B^x) + Cx = D is also a pain?

Capital alphas known constants.

Guess it requires some transcendental meditation? Trouble is, I've got a few to do, different sets of alphas.

**MaxDurango** · October 6th 2008, 07:12 PM

the solution to your problem can be worked out using the natural log. I need a brush up on the rules but basically you take the log of all the numbers. The ones that are multipled turn to addition and the exponent comes down....

(ln 0.72) + (n x ln(1.04)) [not sure how to handle the 0.28] = n x ln 1.035568

i just dont remember the rules surroundint he 0.28,. i dont think you add it at that point (once you take the ln) but you can look that up.

I hit this forum because i am looking at stocks with different yeilds and I know their lines cross when graphed, just trying to figure out where. In my case the "time" is the exponent and its easily solved the way i showed here.... EXCEPT.. I am stuck on basic algebra...... once I get to something like T x A + B = T * C [the t is the unknown and abc are known numbers) I completely spaced on how to solve for T. LOL.

Thread: An equation with unknown exponent

LinkBack

Thread Tools

Display

An equation with unknown exponent

I guess A(B^x) + Cx = D is also a pain?

solve using logs

Similar Math Help Forum Discussions

Matrix equation with unknown coefficents

Solve For the Unknown (ln equation)

1 unknown equation solver

1 equation 1 unknown but still killing my brain

Simultaneous equation 2 unknown vairables

Search Tags