exponential function and inequality

Question

I have a hard time solving this one. I'm sure there is trick that should be used but if so, I can't spot it.

$$(3\cdot4^{-x+2}-48)\cdot(2^x-16)\leqslant0$$

Here is what I get but I'm anything but confident about this:

$$3\cdot(2^{-2x+4}-16)\cdot(2^x-16)\leqslant0$$ $$(2^{-2x+4}-2^4)\cdot(2^x-2^4)\leqslant0$$ $$2^{-x+4}-2^{x+4}-2^{-2x+8}+2^8\leqslant0$$ $$2^{-x+4}+2^8\leqslant2^{x+4}+2^{-2x+8}$$

So far, I'm already not 100% sure but then, I'm not sure at all:

$$(-x+4)\cdot \ln(2)+8\cdot \ln(2)\leqslant(x+4)\cdot \ln(2)+(-2x+8)\cdot \ln(2)$$

This is nonsense, can someone correct me please? Thanks.

Answer 1

To expound on the previous problem (I can't comment), the interval should be

$$(-\infty,0]\cup[4,\infty)$$

For if

$$(1-t)(t-16)\le0$$

then

$$(t-1)(t-16)\ge0$$

So

$$(2^x-1)(2^x-16)\ge 0$$

Note that if $x=2$, we have

$$(4-1)(4-16)=-36$$

so $2$ can not be a solution

Answer 2

If we put $t=2^x>0$, the inequation becomes

$$48(\frac{1}{t^2}-1)(t-16)\le 0$$

which equivalent to $$(1-t)(1+t)(t-16)\le 0$$

or $$(1-t)(t-16)\le 0$$

thus $$2^x\le 1 \; or \;2^x \ge2^4$$ so, the answer is $$x\le 0 \;\; or \;\; x \ge 4$$