<s>[INST] 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.
 [/INST] $$1-3^x+5^x-7^x = 0$$
$$\rightarrow0\geq 1-3^x=7^x-5^x \geq 0 $$
$$\rightarrow0= 1-3^x=7^x-5^x = 0 $$
$$\rightarrow x=0$$
EDIT
If $x<0$ we'd get $$\rightarrow 0\leq 1-3^x=7^x-5^x \leq 0 $$
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