Solve for $x : A^x- B ^x= C $ How to approach this kind of equation Solve for $
x $in the below equation. I have an equation in the below format
$ A^x- B^x= C $
How to approach this kind of equation and solve for $x$. 
I took log on both sides and it doesn't work out for me ..
Any suggestions
 A: If we set $y = A^x$, we can rewrite this as
$$
y - y^{\log_A(B)} = C
$$
So solving your equation (i.e. getting an exact result) is equivalent to solving this one.
Of course, if $C = 0$, then we can always solve this.
Otherwise: If $\log_A(B)$ is an integer, then this is a polynomial, which may not always be "solvable" in a satisfying sense.  Otherwise, we have even less of a chance of getting an exact solution.
So, for arbitrary $A,B,$ and $C$: if you want to solve this, you're better off with some kind of numerical approximation method.
A: In the most general case, the root of equations such $$f(x)= A^x- B ^x- C=0$$ cannot be expressed explicitly and numerical methods (such as Newton) should be required.
Let us admit that $A,B,C$ are real and positive numbers and $A\gt B$. Function $f(x)$ can be very stiff but $$g(x)=x\log(A)-\log(C+B^x)$$ can be almost a straight line which is good for the solver.
For illustration purposes, let us use $A=7, B=5, C=123456789$. If you plot the function, for sure you will notice that the root is somewhere close to $x_0=10$. So, let us try Newton method with function $f(x)$; this will generate the following iterates $$x_1=9.720476380$$ $$x_2=9.609545440$$ $$x_3=9.595611384$$ $$x_4=9.595417884$$ $$x_5=9.595417847$$ which is the solution for ten significant figures.
Let us do the same using function $g(x)$; this will generate the following iterates $$x_1=9.588841160$$ $$x_2=9.595416723$$ $$x_3=9.595417847$$ which is incredibly must faster.
What could be amazing to you is that, being very lazy and starting at $x_0=0$, the first iterate of Newton method would have been $x_1=9.57465$ already quite close to the solution.
For the general case, you could safely start iterating at $$x_1=\frac{(C+1) \log (C+1)}{(C+1) \log (A)-\log (B)}$$ which is the first iterate of Newton method applied to function $g(x)$ starting at $x_0=0$.
