An exponential number algebra problem

[Edge5]

Let a,b,c and n are real numbers.a-b = C
I want to get rid of a,b and find the following expression in terms of C and n. How can I do that?
(aⁿ-bⁿ)= ? (in terms of C and n)

Thank you.

 

[mfb]

You can't.
Simple example: a=2, b=1 and a=1, b=0 both lead to C=1, but a²-b² is different for the two cases.

You can rewrite aⁿ-bⁿ to have a factor of (a-b) but you won't get rid of a and b completely.

What is the context of this question?

 

[Edge5]

You can't.
Simple example: a=2, b=1 and a=1, b=0 both lead to C=1, but a²-b² is different for the two cases.

You can rewrite aⁿ-bⁿ to have a factor of (a-b) but you won't get rid of a and b completely.

What is the context of this question?

I have a resonator and the resonance frequency (w) of it is given by w=A(L^-1.5) where A is a constant and L is the length. When I apply a force the resonance frequency changes because length of the beam changes due to deformation. I need to find the change in resonance frequency as a function of change in length.
That's why I said w_initial = AL_initial^-1.5
and AL_final^-1.5
L_final-L_initial = (L_initial.F_tensile)/(EA_crossection)
Where E is the young modulus and A is the area.
In my question a and b were L_initial and L_final
n was -1.5
C was (L_initial.F_tensile)/(EA_crossection)

 

[BvU]

So where are the $a$ and $b$ in your story ?

 

[Edge5]

So where are the $a$ and $b$ in your story ?

In my question a and b were Linitial and Lfinal

 

[BvU]

So take logarithms ! $$\ln\omega = \ln A - 1.5\ln L$$

 

[Edge5]

So take logarithms ! $$\ln\omega = \ln A - 1.5\ln L$$

I will try thanks