Matematické Fórum

stuart clark · 17. 05. 2011 04:42

For what value of $a$ does the equation $2(log_{3}x)^2-\left|log_{3}x\right|+a=0$ have $4$ solution

byk7 · 17. 05. 2011 16:51

First, did you mean four real solutions?

We know, that the identity $|A|^2=A^2$ works for every real $A$ . So I choose substitution $t:=\left|\log_3(x)\right|.$ Then the equation will be $2t^2-t+a=0.$ Discriminant of this equation is $\textrm{D}=(-1)^2-4\cdot2\cdot a=1-8a.$ The equation will have two (real) solutions, if $a\in\bigg(-\infty,\frac18\bigg)$ (for $a>1/8$ is $\rm D$ negative and there will be only one root for $a=1/8$ ). Solution of equation is $t=\frac14$1\pm\sqrt{1-8a}$.$ Now turn back to the substitution: $\left|\log_3(x)\right|=\frac14$1\pm\sqrt{1-8a}$\Rightarrow\log_3(x)=\pm\frac14$1\pm\sqrt{1-8a}$\Rightarrow\color{blue}x=3^{\pm$1\pm\sqrt{1-8a}$/4}\color{black}.$

The equation has four solution for every $a\in$-\infty,\frac18$$ .

E.g... Zobrazit/skrýt!

stuart clark · 23. 05. 2011 09:23

Thanks byk7.

But I have got $a\in\left(0,\frac{1}{8}\right)$

musixx · 23. 05. 2011 13:02 — Editoval musixx (23. 05. 2011 13:11)

↑ byk7: If a discriminant is nonzero, then we have two distinct solutions of the quadratic equation, let's say a and b. But your construction assumes that all four numbers a,b,-a,-b are mutually distinct. You should extend your answer.

Autor skryl část textu. Zobrazit/skrýt!

halogan · 23. 05. 2011 13:14

↑ byk7:

We have to ensure that both $t$ s are positive (just plot the function $x \to |\log_3 x|$ ) in order to have 4 distinct roots.

From that we get that $a$ must be greater that zero.

byk7 · 23. 05. 2011 13:25

↑ stuart clark:
↑ halogan:
↑ musixx:

Of course, I'm sorry for the mistake.

Matematické Fórum

#1 17. 05. 2011 04:42

4 solution

#2 17. 05. 2011 16:51

Re: 4 solution

#3 23. 05. 2011 09:23

Re: 4 solution

#4 23. 05. 2011 13:02 — Editoval musixx (23. 05. 2011 13:11)

Re: 4 solution

#5 23. 05. 2011 13:14

Re: 4 solution

#6 23. 05. 2011 13:25

Re: 4 solution

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