TITLE: Difficult exponential equation $6^x-3^x-2^x-1=0$
QUESTION [8 upvotes]: $$6^x-3^x-2^x-1=0$$I don't have any idea how to solve this equation. I don't know how to show that the function $f(x)=6^x-3^x-2^x-1$ is strictly increasing (if it's the case). The derivative doesn't show me anything. I tried also to use the Lagrange theorem(like in this equation for example: $2^x+9^x=5^x+6^x$), but it's not the case!
But I need a solution without calculus. Please help!
Edit: In my country the theorem that states roughly this $f'(c)=\dfrac{f(b)-f(a)}{b-a}, $where $f:[a,b]\to\mathbb{R} $ is called Lagrange theorem

REPLY [2 votes]: I quite liked your question so thought this might be of assistance next time you face a question similar to this one you can employ some mathematical software to help for e.g. Maple. You can rewrite the equation to look like this
$$ f(x) = g(x) $$
where $f(x)= 6^x - 3^x -2^x$ and $g(x)=1$ is a constant function. So it's an intersection problem. The graph is below.

Clearly, they intersect at $x=1$ or you can use the following code to be $100\%$ sure 
fsolve({f(x) = g(x)}, {x}, x = -1 .. 2)
where $f(x)$ and $g(x)$ are defined as above.