The total number of solutions (real) of equation: $2^x+3^x+4^x-5^x=0 ?$ The total number of solutions (real) of equation: $2^x+3^x+4^x-5^x=0 ?$
I have no idea how to solve this problem. Can someone point me in the right direction?

$$2^x+3^x+4^x=5^x$$
$${\left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x} =1$$
$$y={\left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x}$$ 
$$y=1$$
Since the function is always decreasing . It will meet $y=1$ only once .
Graph will make things more easier .
