I am currently practising for my maths exam and I have come across this problem, which I can partially do:
$$\log^2_{1/2} x<4+\log_{1/2} x $$
After I put everything on the left hand side I get something like this: $$\log^2_{1/2} x - \log_{1/2} x - 4< 0$$
Then, I simply substitute $t$ as $\log_{1/2} x$: $$t^2 - t - 4 < 0$$
The roots of this quadratic formula are: $x_1 = \frac{1-\sqrt{17}}{2}$ and $x_2 = \frac{1+\sqrt{17}}{2}$, so the formula looks like this: $$ \left(t-\left(\frac{1-\sqrt{17}}{2}\right)\right)\left(t-\left(\frac{1+\sqrt{17}}{2}\right)\right) < 0$$
Here, the interval of $t$ is: $\left(\frac{1-\sqrt{17}}{2}, \frac{1+\sqrt{17}}{2}\right)$.
But here is the problem I get, I don't know what to do with this. All I know is that I somehow need to switch from $t$ to $\log_{1/2} x$ and then find the $x$. I need your help guys and I am desperate. Thanks in advance and have a nice rest of the weekend!
You found that $$ \frac{1 - \sqrt{17}}{2} < t < \frac{1 + \sqrt{17}}{2}$$
Since, we defined $t=\log_{\frac{1}{2}}(x)$
$$\frac{1-\sqrt{17}}{2} < \log_{\frac{1}{2}}(x) < \frac{1+\sqrt{17}}{2}$$
We want to raise all three sides to the power of $\frac{1}{2}$ to get rid of the $\log{x}$ and get an inequality only in terms of $x$. However, here is where we need to be careful.
The rule for exponential inequalities state that if $a>1$ and $x>y$, then $a^x > a^y$. However, if $0<a<1$, then $a^x < a^y$. This is due to the fact that the exponential function $f(x) = a^x$ is strictly increasing (i.e. as $x$ increases, $f(x)$ consistently increases) for $a>1$, and strictly decreasing (i.e. as $x$ increases, $f(x)$ consistently decreases) for $0<a<1$.
The base of the logarithm $\frac{1}{2}$ is between $0$ and $1$, so when we raise all three sides to the power of $\frac{1}{2}$ the inequality signs flip:
$$\left(\frac{1}{2}\right)^{\frac{1-\sqrt{17}}{2}} > \left(\frac{1}{2}\right)^{\log_{\frac{1}{2}}(x)} > \left(\frac{1}{2}\right)^{\frac{1+\sqrt{17}}{2}}$$
$$\left(\frac{1}{2}\right)^{\frac{1-\sqrt{17}}{2}} > x > \left(\frac{1}{2}\right)^{\frac{1+\sqrt{17}}{2}}$$
$$\left(\frac{1}{2}\right)^{\frac{1+\sqrt{17}}{2}} < x < \left(\frac{1}{2}\right)^{\frac{1-\sqrt{17}}{2}}$$