Logarithmic equation $\log_{3x}(3/x)+(\log_3(x))^2=1$.

Question

The equation is $$\log_{3x}\frac{3}{x}+\log^2_3x=1$$ I tried to solve it like this $$\log_{3x}\frac{x^{-1}}{3^{-1}}+\log^2_3x=1$$ $$\log_{3x}\left(\frac{x}{3}\right)^{-1}+\log^2_3x=\log_33$$ $$-\log_{3x}\frac{x}{3}+\log^2_3x=\log_33$$ $$-\log_{3x}x-\log_{3x}3+\log^2_3x=\log_33$$ I don't have any other ideas.

Answer 1

Hint:

Express everything in base $3$ and make some substitution like $t= \log_3x$. Also use addition theorem for log.

For an example: $$\log_{3x}x = {\log_3x\over \log_3(3x)} = {\log_3x\over \log_33-\log_3x}={t\over 1-t}$$

So, you got $${1-t\over 1+t} = 1-t^2= (1-t)(1+t)$$

If we multiply by $1+t$ we get after some rearrangement $t^3+t^2-2t=0$. This we can factor: $t(t^2+t-2)= t(t+2)(t-1)=0$, so $t \in \{-2,0,1\}$. Can you finish now?

Answer 2

Perhaps it is simpler to rewrite $$ 1 = \log_{3x}(3/x)+(\log_3(x))^2 = \frac{\ln 3 - \ln x}{\ln 3 + \ln x} + \left(\frac{\ln x}{\ln 3}\right)^2 $$ and substitute $u = \ln x$.

Answer 3

Hint: writing your equation in the form $$\log_{3x}{3}-\log_{3x}{x}+\left(\frac{\ln(x)}{\ln(3)}\right)^2=1$$ And this is $$\frac{\ln(3)}{\ln(3)+\ln(x)}-\frac{\ln(x)}{\ln(3)+\ln(x)}+\frac{(\ln(x))^2}{(\ln(3))^2}=1$$ Can you finish? Substitute $t=\ln(x)$ Simplifying and factorizing we get $$t(3\ln(3)+t)(\ln(3)-t)=0$$ Multiplying by $$\ln(3)+t,\ln^2(3)$$ we get $$(\ln(3)-t)\ln^2(3)+t^2(\ln(3)+t)=\ln^2(3)(\ln(3)+t)$$

Answer 4

HINT

We have

$$\log_{3x}\frac{3}{x}+\log^2_3x=1$$

$$\log_{3x}3-\log_{3x}x+\log^2_3x=1$$

$$\log_{3x}3x-2\log_{3x}x+\log^2_3x=1$$

$$1-2\log_{3x}x+\log^2_3x=1$$

$$-2\log_{3x}x+(\log_3 x)^2=0$$

$$-2\frac1{\log_{x}3x}+\left(\frac{1}{\log_x 3}\right)^2=0$$