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The inequality for x is:

$(\log_3 x)^2 \lt \log_9( x^4)$

I plotted the graphs to find as answer: $1\lt x \lt 9$

My approach:

In my attempt to solve it, I used the logarithm properties to equal the bases:

$\log_3x \cdot \log_3x \lt 4\log_9x$

$\log_3x \cdot \log_3x \lt 4\log_3x^{1/2}$

$\log_3x \cdot \log_3x \lt 2\log_3x$

Then I changed variables, letting $u = \log_3x$, thus:

$u^2 \lt 2u$

$u^2 - 2u \lt0$

$u(u-2) \lt 0$

$u < 0$ or $u < 2$

Replacing the prior variable:

$\log_3x\lt0$ and $\log_3x\lt2$

Which gives: $x\lt 3^0$ and $x\lt3^2$:

$x<0$ and $x\lt9$.

Thats not the answer. I would appreciate if someone could point out my mistake.

Edit: I made a mistake when typing the first equations. The right Log was supposed to have the base 9, not 3, as initially was there for some of you to solve. Now it's fixed.

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    $\begingroup$ Unfortunately, there are lots of mistakes. In your first line, you change to a base $9$ logarithm and erase the argument. In your second line, you have an exponent $1/2$ on the right, but not on the left. In your third line, you've taken a square root of only part of the expression. Later on you say that $u(u - 2) < 0$ implies that $u < 0$ or $u < 2$, which is incorrect. Shortly after that, you change 'or' to 'and.' Finally, $3^0 = 1$, not $0$. $\endgroup$ – user296602 May 21 '18 at 22:19
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    $\begingroup$ If in fact the inequality should have read $$(\log_3 x)^2<\log_9(x^4)$$ your u-inequality $u(u-2)<0$ is in fact correct. This implies $0<u<2$ and after applying $3^t$ on all sides we have $1<x<9$ as expected. $\endgroup$ – String May 21 '18 at 22:29
  • $\begingroup$ I'm sorry for the misunderstanding. I've mistakenly switched the base 9 in the left logarithm to a 3 $\endgroup$ – Lucas Lemos May 21 '18 at 22:31
  • $\begingroup$ But as @T.Bongers points out, you should do a thorough cleanup of what you have so far. $\endgroup$ – String May 21 '18 at 22:32
  • $\begingroup$ @LucasLemos: Then my comment applies and your mistakes are merely reversing one inequality $u<0$ in stead of $0<u$ and applying $3^t$ incorrectly to $t=0$, since $3^0=1\neq 0$. $\endgroup$ – String May 21 '18 at 22:33
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Your work is good up to $u(u-2)<0$. Then you mistakenly conclude that $u<0$ or $u<2$ instead of the correct $0<u<2$.

Once you realize this, you have immediately $0<\log_3x<2$ that becomes $1<x<3^2$.

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Put $z = \log_3(x)$. You now have $z^2 < 4z$. Now solve that inequality and unwind it.

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  • $\begingroup$ FWIW, the asker does make this substitution and arrives at that inequality - these are both contained in the question. But there are so many other things wrong that they get a rather incorrect result. $\endgroup$ – user296602 May 21 '18 at 22:28
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I am not sure why you think you need to "equal the bases"? Everthing is already base 3.

But you do want to use the rules of exponents... and the LHS of your next line is good.

$(\log_3 x)^2 < \log_3 x^4\\ (\log_3 x)(\log_3 x) < 4\log_3 x\\ (\log_3 x)(\log_3 x - 4) < 0$

Either...

$\log_3 x > 0$ and $\log_3 x < 4$

or

$\log_3 x < 0$ and $\log_3 x > 4$

But that is not possible.

$\log_3 x > 0$ and $\log_3 x < 4$

$x > 3^0$ and $x < 3^4$

$1<x<81$

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