1
$\begingroup$

If $\,2 \log_x (a)+ \log_{ax} (a) + 3\log_{a2x}, (a)=0,\,$ then the equation has how many real roots?

The above problem is a quadratic equation problem.

$\endgroup$
  • $\begingroup$ I edited the posting, can you verify that the equation is what you intended it to be? especially the part about $a2x$, is it $a^2x$ or something else? $\endgroup$ – Siong Thye Goh Jul 25 '16 at 17:38
0
$\begingroup$

The big thing to keep in mind here is the change of base formula:

$$\log_p(x) = \frac{\log_q(x)}{\log_q(p)}$$

Then $$\log_{ax}(a) = \frac{\log_x(a)}{\log_x(ax)} = \frac{\log_x(a)}{\log_x(a)+\log_x(x)} = \frac{\log_x(a)}{\log_x(a)+1}$$

and

$$\log_{a^2x}(a) = \frac{\log_x(a)}{2\log_x(a) + 1}.$$

This gives you the following:

$$2 \log_x a+ \log_{ax} a + 3\log_{a^2x} a$$

$$=2 \log_x a+ \frac{\log_x(a)}{\log_x(a)+1} + 3\frac{\log_x(a)}{2\log_x(a) + 1}$$

Subsituting $u=\log_x(a)$ yields,

$$2u + \frac{u}{u+1} + \frac{3u}{2u+1} = 0$$

Try taking it from here.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

use $\log_{a}b=\frac{1}{\log_{b}a}$ and then take $1/(log_{a}x)=d$ to get the answer .(A polynomial in d)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Sir can post the full answer please $\endgroup$ – Anirudh Raghavan S Jul 25 '16 at 17:44
0
$\begingroup$

I would replace $\log_x{a}$ (and any other 'log' term in fact similarly) by $\dfrac{\log_{10}{a}}{\log_{10}{x}}$ everywhere, keeping in mind that $a>0$ and $a\ne1$. Now, I can rewrite your equation:

$2\log_x{a}+ \log_{ax}{a} + 3\log_{a^2x}{a}=0$

as -

$2\dfrac{\log_{10}{a}}{\log_{10}{x}}+\dfrac{\log_{10}{a}}{\log_{10}{ax}}+3\dfrac{\log_{10}{a}}{\log_{10}{a^2x}}=0$

Now, we can use the addition property of logarithms (i.e. the fact that $\log_b{xy}=\log_b{x}+\log_b{y}$) to rewrite the above expression as -

$2\dfrac{\log_{10}{a}}{\log_{10}{x}}+\dfrac{\log_{10}{a}}{\log_{10}{a}+\log_{10}{x}}+3\dfrac{\log_{10}{a}}{\log_{10}{a}+\log_{10}{a}+\log_{10}{x}}=0$

Let $\log_{10}{a}=A$ and $\log_{10}{x}=t$, so that the equation now becomes -

$2\dfrac{A}{t}+\dfrac{A}{A+t}+3\dfrac{A}{2A+t}=0$

You can transform this into a quadratic by taking L.C.M. and then solving as usual (keeping in mind that $t\ne0,-A,-2A$). But it's easy to see that $t=-A/2$ is a root of the above equation. That will make your task slightly easier. The quadratic that you would get after taking L.C.M. of the fractions on the left hand side of the above equation would be -

$6At^2 + 11A^2t + 4A^3=0$

As $a\ne1$ and therefore, $A=\log_{10}{a}\ne0$, we can divide by $A$ to get -

$6t^2+11At+4A^2=0$

Now, sum of roots of the above quadratic would be $-11A/6 = -A/2+t_2$, where $t_2$ is the other root. Thus, the two roots will be

$t=-A/2$ and $t=-4A/3$, or in terms of $x$ we have -

$x=a^{-1/2}$ and $x=a^{-4/3}$. Thus, we have two real roots.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.