Inequality $\frac{8}{3^x-2}\geq 3^x$.

Question

Give the sum of the integers that solve the inequality $$\frac{8}{3^x-2}\geq 3^x.$$

Since $f(x)=3^x>0, \ \forall x\in\mathbb{R},$ we can start by doing the substitution $t=3^x$ and get the inequality $$\frac{8}{t-2}-t\geq 0 \Longleftrightarrow f(t) = -\frac{(t+2)(t-4)}{t-2}\geq 0.$$

Looking at the signs of the f(t) we can generate the following table:

$$\begin{array}{|l|cr} t & & -2 & & 2 & & 4 &\\ \hline -(t+2)(t-4) & + & 0 & + & 0 & + & 0 & - & \\ t-2 & - & - &-& 0 & - & + & + & \\ f(t) & - & 0 & - &\varnothing & - & 0 & - & \end{array}$$

So it seems that $f(t)\geq 0$ only at $t_1=-2$ and $t_2=4$. Going back to $x$, we have to discard $t_1$. The only solution seems to be $3^x = 4 \Leftrightarrow x = 2\ln{2}/\ln{3}.$

Seriously, I have no idea what is going on here. Am I being unreasonable for doing that substitution? The answer to this problem is 1.

Answer 1

Let $3^x=t$.

Hence, $t>0$ and we need to solve $$\frac{8}{t-2}\geq t$$ or $$\frac{(t+2)(4-t)}{t-2}\geq0$$ or $$2<t\leq4$$ or $$2<3^x\leq4$$ or $$\ln_32<x\leq2\log_32,$$ which gives $x=1$.

Answer 2

Note that $(3^x-2)^2>0$.

\begin{align*} \frac{8(3^x-2)^2}{3^x-2}&\ge 3^x(3^x-2)^2\\ 8(3^x-2)&\ge 3^x(3^x-2)^2\\ (3^x-2)[3^x(3^x-2)-8]&\le 0\\ (3^x-2)[(3^x)^2-2(3^x)-8]&\le 0\\ (3^x-2)(3^x-4)(3^x+2)&\le 0\\ (3^x-2)(3^x-4)&\le 0 \qquad\qquad (\text{Note that }3^x+2>0)\\ 2\le 3^x&\le 4\\ x&=1 \end{align*}