Is this a valid method of solving $3^x + 2^x = 35$? I want to know if this is a valid method of solving this equation:
$3^x + 2^x = 35$
$3^x+2^x = (7)(5)$
$3^x+2^x = (3+2^2)(2+3)$
$3^x+2^x = 3^2 + 2^3 + 18$
$3^x+2^x = 9 + 18 + 2^3 $
$3^x+2^x = 27 + 2^3 $
$3^x+2^x = 3^3 + 2^3 $
And now comes my problem. Is it correct to say that this last equation implies $x=3$? I don't think I can just compare terms just like it was a polynomial...
If this is not correct, is there a way of solving this equation algebraically?
 A: Yes, it is a valid method of producing one solution. An additional argument may be called for to show that there are no other solutions. For example, you could argue that $3^x+2^x$ is a one-to-one function.
How do you call the method you used, compared to "algebraically"?
In general, there are no "instant" ways to solve such equations. The fact you were able to solve it in such a nice way is very special. Consider, for example, the equation

$$3^x+2^x=36.21923341032$$

We know that there is a unique solution, (using the mean value theorem, for instance), but there is no way to generate it in closed form, or any other form, except from numerical approximation.
A: $3$ it's an unique root because $f(x)=3^x+2^x$ is an increasing function and $f(3)=35$.
A: The claim is equivalent to 
$$
f(x)=f(3)\implies x=3
$$
where $f(x)=3^x+2^x$. This is true because $f$ is injective (since it is strictly increasing for example).
A: As others have noted, $x=3$ is the only solution in the real numbers.  However, there are infinitely many complex solutions.  For example, there is a solution near $3.40258654642902+5.90727700642090 i$.
