Given the following function how many solutions to $f(x)=0$ are there? I have the following function:
$f: \mathbb{R} \rightarrow \mathbb{R}$ 
$f(x)=9^x-5^x-4^x$
And I have to find the number of real soltutions (so not necessarily the solutions themselves, just how many are there) for $f(x)=0$ and $f(x)-2 \sqrt{20^x}=0$.
For $f(x)=0$ this is what I did:
$9^x-5^x-4^x=0$
I divided by $9^x$,
$1- \bigg (\dfrac{5}{9} \bigg)^x - \bigg (\dfrac{4}{9} \bigg)^x = 0$
$\bigg (\dfrac{5}{9} \bigg)^x + \bigg (\dfrac{4}{9} \bigg)^x = 1$
Since the left-hand side is the sum of two strictly decreasing functions, I conculded that the left-hand side is strictly decreasing ($1$). So the equation can have at most $1$ solution. By pure guessing, I found that $x=1$ is a solution to the equation and because of ($1$) it is the only solution. So, $f(x)=0$ has only one solution ($x=1$ to be precise, but this is not necessary). I think I got this right, but if someone finds a mistake, please let me know.
My real trouble is at the second part of the problem, where I have to find the number of solutions for:
$9^x-5^x-4^x - 2 \sqrt{20^x} = 0$
I don't see how should I approach this. Any help will be appreciated.

Yes we have that
$$f(x)=9^x-5^x-4^x=0 \iff \left(\frac49\right)^x+\left(\frac59\right)^x=1$$
and since 
$$ g(x)=\left(\frac49\right)^x+\left(\frac59\right)^x\implies g'(x)=-\left(\frac49\right)^x\log \left(\frac94\right)-\left(\frac95\right)^x\log \left(\frac95\right)<0$$
$f(x)$ is strictly decreasing and


*

*$\lim_{x\to \infty} f(x)=0$

*$\lim_{x\to -\infty} f(x)=\infty$
by IVT we have exactly one solution for $f(x)=1$.
For the second equation we can use that
$$f(x)=9^x-5^x-4^x - 2 \sqrt{20^x}=(3^x)^2-((\sqrt 5)^x+(\sqrt 4)^x)^2=\\=(3^x+2^x+(\sqrt 5)^x)(3^x-2^x-(\sqrt 5)^x)=0$$
that is $3^x-2^x-(\sqrt 5)^x=0$ which can be studied in the same way.
