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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 $1$.

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  • $\begingroup$ 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. $\endgroup$
    – J.G.
    Commented Apr 2 at 15:16
  • $\begingroup$ Ok but how do the logs disappear? $\endgroup$
    – koiboi
    Commented Apr 2 at 15:17
  • $\begingroup$ By applying $f^{-1}$ to the original inequality. $\endgroup$
    – J.G.
    Commented Apr 2 at 15:18
  • $\begingroup$ So he raises the bases to the logs itself doesnt he? $\endgroup$
    – koiboi
    Commented Apr 2 at 15:21
  • $\begingroup$ $a < \log_b(y) < c$ 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) $\endgroup$
    – Henry
    Commented Apr 2 at 15:30

2 Answers 2

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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 $n\geq 1$

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    $\begingroup$ 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. $\endgroup$ Commented Apr 2 at 15:48
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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 : $$log_{A}{(x^2 - 7x + 13)} < 0$$

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

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