Solve exponential equation $3^x= 2^x+2$ How do we solve this? I can't think of an easy way.. Is there any way to solve it without using newton's method or other approximations?
$3^x=2^x+2$
 A: First of all, notice that the Newton-Raphson Method will almost never solve an equation. It will give you some very good numerical approximations, but it will almost never find the exact answer.
There is no algebraic way of solving the equation $3^x=2^x+2$. There are, of course, may ways of finding numerical approximations. You could use:


*

*Interval Bisection

*Linear Interpolation
I am unsure as to why you don't want to use the N-R-Method. It is a very good method, especially when solving equations like the one you have given. 
In my humble opinion, the best way to approximate a solution is through the N-R-Method. I get
$$x \approx 1.39495460217075$$
A: Not that I know of either.  There is clearly one positive root, which is between 1 and 2.  Alpha finds it quite close to $1.4$
A: As said, there is no analytical solution to the equation and Newton would be the best candidate for this problem.
If you really do not want to use Newton, you could rewrite the equation as $$x_{n+1}=\frac{\log \left(2^{x_n}+2\right)}{\log (3)}$$ Starting at $x_0=0$, the  successive iterates would be :$1.00000$, $1.26186$, $1.34821$, $1.37832$, $1.38901$, $1.39283$, $1.39419$, $1.39468$, $1.39486$, $1.39492$, $1.39494$, $1.39495$ which is the solution for six significant figures.
Just by curiosity, let us see what Newton would give writing the function as $$f(x)=x-\frac{\log \left(2^{x}+2\right)}{\log (3)}$$ Starting at $x_0=0$, the  successive iterates would be :$1.26632$, $1.39359$, $1.39495$. The same solution is obtained much faster.
