Solving an exponential inequality

Question

$$(0{,}25)^{3-0{,}5x^2}\leq8$$

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.

$$\begin{aligned} (0{,}25)^{3-0{,}5x^2} &\leq 32\,(0{,}25) \\ 3 &\leq 0{,}5\,x^2 \\ x &\geq 2{,}45 \end{aligned}$$

Answer 1

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 $-3\le x\le 3$.

Answer 2

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 $.

Answer 3

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 $[-3,3]$.