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$\log_3x^3 + {3\over \log_3x} =4$

Ok, the way the computer has put it makes it look weird.But it is :log to base 3 of "x" to power 3 plus 3 divided by log to base 3 of x is equals to "4". This question was given to me by a guy from Kenya and I asked our math teacher and it made him scratch his head.So we never reached a conclusion.This was what I was able to do so far:

Let $\log_3 x= u$

$\log_3 x^3 + {3 \over \log_3 x} = 4$

$3 \cdot \log_3 x + {3 \over \log_3 x} = 4$

$3u + 3/u = 4$

$3u^2 + 3 = 4u$

$3u^2 - 4u + 3 = 0 $

Note that discriminant will be $(-4)^2 - 4(3)(3) = 16 - 36 = -20$, which is $< 0$. This means that there is no real solution to the derived equation, i.e., there is no real value of $x$ which will fit the original equation.

So what the heck is the answer!?!?

The question had an answer and though I may not remember it, I do know that if you back substituted it, it would have worked.

Is there any other way to do this question apart from the conventional means?

And also I happen to be in highschool so dont explain using concepts I would have no idea about.

Thankyou in advance!

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The equation $$ 3u^2 -4u+3 = 0$$ has no real solutions so your question has no real solutions. As far as i can see, your method is ok so far.

Using the quadratic formula, $$ u = \frac{2}{3}\pm i\frac{\sqrt{5}}{3}$$

Hence, $$ x = 3^{\frac{2\pm i\sqrt{5}}{3}}$$

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