# Question regarding nature of logarithmic equations

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 $$1$$.

• If $f(x)>a$ with $f$ an order-preserving invertible function of inverse $f^{-1}$ for which $f^{-1}(a)$ exists, $x>f^{-1}(a)$. If $f$ is invertible but reverses order, you have to reverse the inequality's direction.
– J.G.
Commented Apr 2 at 15:16
• Ok but how do the logs disappear? Commented Apr 2 at 15:17
• By applying $f^{-1}$ to the original inequality.
– J.G.
Commented Apr 2 at 15:18
• So he raises the bases to the logs itself doesnt he? Commented Apr 2 at 15:21
• $a < \log_b(y) < c$ is equivalent to $b^a<y<b^c$ when $b>1$, and to $b^a >y >b^c$ when $0 < b< 1$, because $\log_b$ is a strictly monotonic function (increasing in the first case and decreasing in the second) Commented Apr 2 at 15:30

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 $$n\geq 1$$

• It is worthwhile to note the reason why the order is inverted: because for any base $0<b<1$, $\log_b$ is a strictly decreasing function, so its inverse function is also strictly decreasing. Also, in your notation $n$ must be strictly greater than $1$. Logarithm with base $1$ is not defined. Commented Apr 2 at 15:48

I am writing this out in great Detail , due to matching the OP level.

(1) We make "$$\log$$" disappear by "raising the Base to the Power" :

$$A \le \log_a (B) \le \log_a (C) \le D$$ $$a ^ A \le a ^ {\log_a (B)} \le a ^ {\log_a (C)} \le a ^ D$$ $$a ^ A \le B \le C \le a ^ D$$

It is similar to adding Constant through-out or multiplying by Positive Constant though-out , ETC.
We are "raising to Power" though-out.

In OP Case : $$-1 \leq \log_9(\frac{x^2}{4}) \leq 1$$ $$9^{-1} \leq 9^{\log_9(\frac{x^2}{4})} \leq 9^{1}$$ $$9^{-1} \leq \frac{x^2}{4} \leq 9^{1}$$

(2) When we have Base $$a$$ less than $$1$$ , we use the "rule" like : $$\log_a(B)=\log_A(B)/\log_A(a)$$ , where the new Base $$A$$ might be more than $$1$$.
When Denominator is less than $$0$$ (negative) , it can be eliminated by interchanging the relation between $$\le$$ & $$\ge$$
That "rule" is like : If $$-a > 0$$ , then $$a < 0$$ & vice-versa

In OP Case : $$log_{1/2}{(x^2 - 7x + 13)} > 0$$ $$log_{A}{(x^2 - 7x + 13)}/log_{A}{1/2} > 0$$ Here , though $$A$$ is arbitrary , it must be larger than $$1$$ : We can take $$A=2$$

$$log_{A}{(x^2 - 7x + 13)}/(-1) > 0$$ $$(-1)log_{A}{(x^2 - 7x + 13)} > 0$$

We then interchange $$>$$ to $$<$$ to eliminate the negation : $$log_{A}{(x^2 - 7x + 13)} < 0$$

$$A^{log_{A}{(x^2 - 7x + 13)}} < A^{0}$$ $$(x^2 - 7x + 13) < 1$$ $$(x^2 - 7x + 12) < 0$$ $$(x - 3x)(x - 4x) < 0$$