# Thread: An equation with unknown exponent

1. ## An equation with unknown exponent

Is is possible to solve this equation?

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

If so, how?

2. 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:

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

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

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

in this case one obvious solution is n = 0

3. Originally Posted by p.numminen
Is is possible to solve this equation?

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

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:

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

4. 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.

5. Originally Posted by p.numminen
Is is possible to solve this equation?

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

If so, how?
As the others have mentioned, an obvious solution is $\displaystyle n=0$

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

$\displaystyle 0.72 + 0.28 = 1$

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

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.

7. ## solve using logs

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.