While reading my textbook's chapter about logarithms and seeing the solved examples I noticed in various places that the author was able to make the $\log$ just disappear in a equation or inequality by making some changes on either side of the inequality or the equation for example:
$$-1 \leq \log_9\left(\frac{x^2}{4}\right) \leq 1$$
changes to
$$9^{-1} \leq \frac{x^2}{4} \leq 9^1.$$
This confuses me how does this happen and another curious thing when doing this "logarithm disappearing" on logs with bases less than $1$ the sign of the inequality changes.
For example in one question the following was done
$$\log_{1/2}(x^2 – 7x + 13) > 0$$
changes to
$$x^2 – 7x + 13 < 1$$
I do not understand how does $1$ just appear on the other side and why does the inequality reverse, could someone explain this to me and my this "log disappearing" is different for bases less than $1$ and greater than $1$.
Best Answer
You can see taking the logarithm as the inverse of exponentiation. So if you have $$-1\leq\log_9\left(\frac{x^2}{4}\right)<1\Rightarrow 9^{-1}\leq9^{\log_9\left(\frac{x^2}{4}\right)}<9$$ which simplifies as your teacher told you. As for the second example the same thing applies: $$\log_{\frac{1}{2}}(x^2-7x+13)>0\Rightarrow x^2-7x+13<\left(\frac{1}{2}\right)^0=1$$ here the $">"$ is inverted because for $n\in (0,1)$ $\log_n(x)$ inverts the order, while it preserves it for $n\geq 1$