Solving the logarithmic inequality $(\log_3 x)^2 \lt \log_9( x^4)$

Question

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.

Answer 1

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 $1<x<3^2$.

Answer 2

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

Answer 3

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 $x < 3^4$

$1<x<81$