I don't have the slightest idea about how to tackle this one. I could change $2\log_x 3$ to $\frac{2}{\log_3 x}$ and deducting that from $\log_3 x$ would give me $\frac{(\log_3x)^2-2}{\log_3x}$, but I don't know how to proceed.

  • $\begingroup$ It looks like you are very close to a quadratic in $\log_3x$... Can you write the full equation resulting from your work thus far? Also, I believe you get $\frac 2{\log_3x}$ as the changed value... $\endgroup$ – abiessu Dec 13 '18 at 13:05
  • $\begingroup$ Thanks! Fixed that... $\endgroup$ – Nameless King Dec 13 '18 at 13:15

$$\log_3 x-2\log_x 3 = 1$$

Here, there is a very good way to manipulate the expression. You have

$$\log_a b = \frac{1}{\log_b a}$$

Applying it here, you get

$$\frac{1}{\log_x 3}-2\log_x 3 = 1$$

Now, let $t = \log_x 3$. You get

$$\frac{1}{t}-2t = 1$$

which results in a quadratic equation. Can you work out the rest?

Of course, your way also works in a similar manner, except you made an error in your work: $2\log_x 3 = \frac{2}{\log_3 x} \color{red}{\neq \frac{1}{2\log_3 x}}$.


If $\log_3x=y\implies x=3^y=?$

we have $$y-\dfrac2y=1\iff0=y^2-y-2=(y-2)(y+1)$$

$\implies y=?$


Write $$\log_3 x - 2\log_x 3 = 1$$ as $$\frac{\log x}{\log 3}-2\frac{ \log 3}{\log x}=1$$ (Taking care that $x\neq 1$)

Take LCM: $$(\log x) ^2 -2(\log 3)^2 =\log x \cdot \log 3$$ Now substitute $$\log x =X \text { and } \log 3=y$$ You get: $$X^2-2y^2=Xy$$ or $$X^2-Xy-2y^2=0$$ Which factorises to:


So, $$X=2y \text{ or } X=-y$$ Substitute $x$ and $y$ back:
$$\log x=2\log 3$$or$$ \log x=-\log 3$$ So, $$\log x= \log 3^2=\log 9$$ or $$\log x=\log 3^{-1}=\log \frac 13$$

Giving you

$$x=9 \text{ or } x=\frac 13$$


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