Prove that $1-3^x+5^x-7^x = 0 \Leftrightarrow x=0$ Problem :
Prove that equation $$1-3^x+5^x-7^x = 0$$
has unique root $x=0$.

Since $\lim_{x\to-\infty}(1-3^x+5^x-7^x) = 1$ and $\lim_{x\to\infty}(1-3^x+5^x-7^x) = -\infty$, 
I tried to show $x \to 1-3^x+5^x-7^x$ is decreasing function, but it wasn't easy to determine its derivative is negative.
Thanks for any help.

Note that $$1-3^x+5^x-7^x = 0 \iff 1+5^x = 3^x+7^x$$
Since the exponential functions involved are increasing functions we have the following inequalities.
$$x>0\implies 3^x+7^x > 1+5^x$$
 and $$x<0\implies 3^x+7^x < 1+5^x$$
Thus the equality happens if and only if  $x=0$ 
