<s>[INST] $|a^{2x}+a^{x+2}-1|\ge 1$ equation for positive a If $|a^{2x}+a^{x+2}-1|\ge 1$ for all values of a(a>0), $\ne 1$. Find the domain of x.
I tried to substitute $a^x=t$ and used the following $|
t^{2}+a^2t-1|\ge 1$ but it is getting complicated.
 [/INST] Assuming that $a, t > 0$,
If $t^2 + a^2 t - 1 < 0$, then
\begin{align}
   |t^2 + a^2 t - 1| &\ge 1 \\
   1 - a^2 t - t^2 &\ge 1 \\
   t^2 + a^2 t &< 0
\end{align}
Which has no positive real solution.
If $t^2 + a^2 t - 1 \ge 0$, then
\begin{align}
   |t^2 + a^2 t - 1| &\ge 1 \\
   t^2 + a^2 t - 1 &\ge 1 \\
   (t + \frac 12 a^2)^2 &\ge 2 + \frac 14 a^4 \\
   t + \frac 12 a^2 &\ge \frac 12\sqrt{8 + a^4} \\
   t &\ge \frac 12\left(\sqrt{8 + a^4} - a^2 \right) \\
   t &\ge \dfrac{4}{\sqrt{8 + a^4} + a^2} \\
   a^x &\ge \dfrac{4}{\sqrt{8 + a^4} + a^2} \\
   x \ln a &\ge \ln 4 -\ln(\sqrt{8 + a^4} + a^2) \\
   x &\ge \dfrac{\ln 4 - \ln(\sqrt{8 + a^4} + a^2)}{\ln a} \\
   &\text{OR} \\
   a^x &\ge \frac 12\left(\sqrt{8 + a^4} - a^2 \right) \\
   x \ln a &\ge \ln{\left(\sqrt{8 + a^4} - a^2 \right)} - \ln 2 \\
   x &\ge \dfrac{\ln{\left(\sqrt{8 + a^4} - a^2 \right)}}{\ln a}
\end{align}
</s>