<s>[INST] $(3^x-2^x)^2+3^x-2^x\leq 5^x$ Solve the following inequality
$$(3^x-2^x)^2+3^x-2^x\leq 5^x$$
I'm having difficulty in solving this inequality because there are multiple exponential with different basis. I thought, however, the replacement:
$$y=3^x-2^x$$
and so:
$$y^2+y\leq 5^x$$
treating $5^x$ as the known term of the inequality of the second degree in y.
What do you think? thank you very much
 [/INST] Your approach will flesh out as follows:
$$\boxed{y^2+y\le5^x}\iff y^2+y-5^x\le0\iff y\in\Big[y_{_1},y_{_2}\Big],\qquad y_{_{1,2}}=\frac{-1\pm\sqrt{5^x+1}}2$$
$$-\sqrt{5^x+1}\le2y+1\le\sqrt{5^x+1}\iff|2y+1|\le|\sqrt{5^x+1}|\iff(2y+1)^2\le5^x+1$$
$$4y^2+4y\le5^x\iff\boxed{y^2+y\le\dfrac{5^x}{4\ \ }}\iff y\in\Big[y'_{_1},y'_{_2}\Big],\qquad y'_{_{1,2}}=\frac{-1\pm\sqrt{\dfrac{5^x}{4\ \ }+1}}2$$
$$4y^2+4y\le\dfrac{5^x}{4\ \ }\iff\boxed{y^2+y\le\dfrac{5^x}{4^2}}\iff y\in\Big[y^"_{_1},y^"_{_2}\Big],\qquad y^"_{_{1,2}}=\frac{-1\pm\sqrt{\frac{5^x}{4^2}+1}}2$$
Traveling down this road, we deduce that $y\in\Big[Y_{_1},Y_{_2}\Big]$, where $Y_{_{1,2}}=\displaystyle\lim_{n\to\infty}\dfrac{-1\pm\sqrt{\frac{5^x}{4^n}+1}}2=\pm1.$
So $-1\le3^x-2^x\le1\iff2^x-1\le3^x\le2^x+1$. Since $\displaystyle\lim_{t\to-\infty}a^t=0\iff x\in(-\infty,x_{_0})$, for some numerically determined value of $x_{_0}$.
</s>