The number of non-negative real roots of $2^{x}-x-1$ are
$ a.)\ 0\\
b.) \ 1 \\
c.)\ 2 \\
d.)\ 3 \\$
I don't have any clue.
I have only learned to solve quadratics and cubic equations ,
i haven't been taught to solve such type of equations where
$x$ is written in power.
I look for a short and simple way.
I would prefer a way $\color{red}{\text{without calculus}}$ unless necessary.
I have studied maths up to $12$th grade .Thanks.
Best Answer
Consider the function $$f(x)=2^x-x-1$$ Compute the derivative $$f'(x)=2^x \log (2)-1$$ The derivative cancels for $$x_*=-\frac{\log (\log (2))}{\log (2)}$$ For this value $$f(x_*)=-1+\frac{1}{\log (2)}+\frac{\log (\log (2))}{\log (2)}\approx -0.0860713$$ The second derivative test would show that this is a minimum. So, two real roots.
By inspection, $x=0$ is a root and $x=1$ another. These are the roots.