Solve the inequality log_3 (tan² x-1) >= log_5 (25)

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Ozma

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Solve the inequality [MATH]\log_3(\tan^2 x-1) \geq \log_5 25[/MATH].

My work: it must be [MATH]\tan^2 x-1>0 \iff -\frac{\pi}{2}+k\pi < x < -\frac{\pi}{4}+k\pi \lor \frac{\pi}{4}+k\pi < x < \frac{\pi}{2}+k\pi[/MATH], with [MATH]k\in\mathbb{Z}[/MATH].
Since [MATH]\log_5 25=2=\log_3 9[/MATH], the inequality is equivalent to

[MATH]\log_3(\tan^2 x-1) \geq \log_3 9 \iff \tan^2 x -1 \geq 9 \iff \tan^2 x \geq 10 \iff \tan x \geq \sqrt 10 \lor \tan x \leq -\sqrt{10}[/MATH]
Let [MATH]\alpha= \arctan \sqrt{10}[/MATH], it is [MATH]\tan x \geq \sqrt 10 \lor \tan x \leq -\sqrt{10} \iff x \geq \alpha +k\pi \lor x \leq -\alpha +k\pi[/MATH]; so, combining this with the condition on the logarithm existence in reals, the solution set I get is

[MATH]-\frac{\pi}{2}+k\pi < x \leq -\alpha+k\pi \lor \alpha+k\pi \leq x < \frac{\pi}{2}+k\pi[/MATH]
However, my textbook has this solution: [MATH]\alpha+k\pi \leq x \leq \pi-\alpha+k\pi[/MATH] with [MATH]\alpha=\arctan \sqrt{10}, \frac{\pi}{3} <\alpha<\frac{\pi}{2}, x \ne \frac{\pi}{2}+k\pi[/MATH]I'm not sure if they are equivalent, or if I'm doing something wrong or if the solution is not correct; I've tried substituting [MATH]\arctan \sqrt{11}[/MATH] (since it is [MATH]\arctan \sqrt{10} < \arctan \sqrt{11}< \frac{\pi}{2}[/MATH]) and the inequality is satisfied, can someone help me understand if I'm doing something wrong or my work is correct? Thanks.
 
Here is a solution of an equivalent form of this question.
I used the change of base theorem: \(\log_3(f(X))=\dfrac{\log(f(X))}{\log(3)}\).
 
Ozma said:
Solve the inequality [MATH]\log_3(\tan^2 x-1) \geq \log_5 25[/MATH].

My work: it must be [MATH]\tan^2 x-1>0 \iff -\frac{\pi}{2}+k\pi < x < -\frac{\pi}{4}+k\pi \lor \frac{\pi}{4}+k\pi < x < \frac{\pi}{2}+k\pi[/MATH], with [MATH]k\in\mathbb{Z}[/MATH].
Since [MATH]\log_5 25=2=\log_3 9[/MATH], the inequality is equivalent to

[MATH]\log_3(\tan^2 x-1) \geq \log_3 9 \iff \tan^2 x -1 \geq 9 \iff \tan^2 x \geq 10 \iff \tan x \geq \sqrt 10 \lor \tan x \leq -\sqrt{10}[/MATH]
Let [MATH]\alpha= \arctan \sqrt{10}[/MATH], it is [MATH]\tan x \geq \sqrt 10 \lor \tan x \leq -\sqrt{10} \iff x \geq \alpha +k\pi \lor x \leq -\alpha +k\pi[/MATH]; so, combining this with the condition on the logarithm existence in reals, the solution set I get is

[MATH]-\frac{\pi}{2}+k\pi < x \leq -\alpha+k\pi \lor \alpha+k\pi \leq x < \frac{\pi}{2}+k\pi[/MATH]
However, my textbook has this solution: [MATH]\alpha+k\pi \leq x \leq \pi-\alpha+k\pi[/MATH] with [MATH]\alpha=\arctan \sqrt{10}, \frac{\pi}{3} <\alpha<\frac{\pi}{2}, x \ne \frac{\pi}{2}+k\pi[/MATH]I'm not sure if they are equivalent, or if I'm doing something wrong or if the solution is not correct; I've tried substituting [MATH]\arctan \sqrt{11}[/MATH] (since it is [MATH]\arctan \sqrt{10} < \arctan \sqrt{11}< \frac{\pi}{2}[/MATH]) and the inequality is satisfied, can someone help me understand if I'm doing something wrong or my work is correct? Thanks.
Click to expand...

Have you tried graphing both answers (and also graphing [MATH]y=\log_3(\tan^2 x-1)[/MATH] to compare with both answers)?

Their answer is

[MATH]\alpha+k\pi \leq x \leq \pi-\alpha+k\pi[/MATH] with [MATH]x \ne \frac{\pi}{2}+k\pi[/MATH]​

This can be rewritten as

[MATH]\alpha+k\pi \leq x \lt \frac{\pi}{2}+k\pi \lor \frac{\pi}{2}+k\pi \lt x \leq \pi-\alpha+k\pi[/MATH]​

Your answer is

[MATH]-\frac{\pi}{2}+k\pi < x \leq -\alpha+k\pi \lor \alpha+k\pi \leq x < \frac{\pi}{2}+k\pi[/MATH]​

Do these agree?
 
Your answer is correct.
First consider the solutions in the range [MATH](-\tfrac{\pi}{2}, \tfrac{\pi}{2})[/MATH][MATH]\tan^2 x≥1, \tan^2 x≥\sqrt{10} \rightarrow \tan^2 x≥\sqrt{10}[/MATH]
so [MATH](-\tfrac{\pi}{2},-\alpha] \cup [\alpha, \tfrac{\pi}{2})[/MATH]so the full solution set: (to use your notation)
[MATH](k\pi-\tfrac{\pi}{2},k\pi-\alpha] \cup [k\pi+\alpha, k\pi+\tfrac{\pi}{2})[/MATH]
[MATH]\bigcup \limits_{k\in \mathbb{Z}} (k\pi-\tfrac{\pi}{2},k\pi-\alpha] \cup [k\pi+\alpha, k\pi+\tfrac{\pi}{2})[/MATH]
 
To compare with your book answer, write out the first two pairs of intervals in your answer:
[MATH](-\tfrac{\pi}{2},-\alpha] [\alpha, \tfrac{\pi}{2}) (\pi-\tfrac{\pi}{2},\pi-\alpha] [\pi +\alpha, \pi +\tfrac{\pi}{2})[/MATH]combining the middle two, you get [MATH][\alpha,\pi-\alpha][/MATH] excluding [MATH]\tfrac{\pi}{2}[/MATH]so the full solution may be written:
[MATH][k\pi+\alpha,k\pi+\pi-\alpha][/MATH] excluding [MATH]k\pi+\tfrac{\pi}{2}[/MATH][MATH]\bigcup \limits_{k\in \mathbb{Z}} [k\pi+\alpha,k\pi+\pi-\alpha] \setminus \{k\pi+\tfrac{\pi}{2}, k\in \mathbb{Z}\} [/MATH]
 
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