Solving an exponential inequality $$(0{,}25)^{3-0{,}5x^2}\leq8$$
Answers given are: $[-3;3]$
Below is where I got with this, I'm pretty sure I took a wrong approach here. Any help at all is appreciated.
$$\begin{aligned}
(0{,}25)^{3-0{,}5x^2} &\leq 32\,(0{,}25) \\
3 &\leq 0{,}5\,x^2 \\
x &\geq 2{,}45
\end{aligned}$$
 A: The numbers $0.25$ and $8$ are both "nice" powers of $2$, which allows for a simple approach.   
The right-hand side is $2^3$. Since $0.25=2^{-2}$, the left-hand side is
$(2^{-2})^{3-0.5x^2}$, which is $2^{x^2-6}$.
So our inequality can be rewritten as
$$2^{x^2-6} \le 2^3,$$
or equivalently, by dividing both sides by $2^3$, as
$$2^{x^2-9} \le 1.$$
This inequality holds precisely if the exponent $x^2-9$ is $\le 0$, that is, when $-3\le x\le 3$.
A: I'd take the logarithm of the first inequality :
$$e^{(3-\frac{1}{2}x^2) \ln(\frac{1}{4})} \le 8$$
becomes 
$$ -(3-\frac{1}{2}x^2) \ln(4) \le \ln(8)$$
As $\ln(8) = \ln(2^3) = 3 \ln(2)$ and $\ln(4) = 2 \ln(2)$, we have :
$$3 -\frac{1}{2}x^2 \ge -\frac{3}{2}$$
Leading to $x^2 \le 9 $.
A: Please forgive me if I use a decimal point instead of a decimal comma; I have a hard time using the comma...
First, I don't see how you got from the first to the second line. 
Second, to go from the second to the third line, you seem to have divided both sides by $0.5$ to get $6\leq x^2$, and then you tried taking the square root. Unfortunately, you seem to have forgotten that $\sqrt{x^2}=|x|$, not $x$. so you should really have gotten $\sqrt{6}\leq |x|$, which would have let to $x\geq \sqrt{6}$ or $x\leq -\sqrt{6}$. 
Okay, let's take it from the top. First, let us "bring down" the exponent by taking logarithms. Since the logarithm is a stricitly increasing function, $0\lt a\leq b$ holds if and only if $\ln(a)\leq \ln(b)$ holds. So
$$\begin{align*}
(0.25)^{3-0.5x^2} &\leq 8\\
(3-0.5x^2) \ln(0.25) &\leq \ln(8)\\
-0.5x^2\ln\left(\frac{1}{4}\right) &\leq \ln(8) -3\ln\left(\frac{1}{4}\right)\\
-0.5x^2(-\ln(4)) &\leq \ln(8) + 3\ln(4)\\
x^2\ln(2) &\leq 3\ln(2)+6\ln(2)\\
x^2\ln(2) &\leq 9\ln(2)\\
x^2 &\leq 9.
\end{align*}$$
The last inequality because $\ln(2)\gt 0$, so dividing through by $\ln(2)$ does not affect the inequality sign.
Now we can take square roots on both sides and we get
$$|x|=\sqrt{x^2} \leq 3$$
and this is equivalent to $-3\leq x\leq 3$; i.e., to the solution set being $[-3,3]$.
