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Find the value of $x$ such that $(3-\log_3x)\log _{3x}3=1$.

Is there another way to solve other than this attempt? My attempt,

$(3-\log_3x)\log _{3x}3=1$

$\frac{\log(3)\left(3-\frac{\log (x)}{\log (3)}\right)}{\log (3x)}=1$

$\log (3)\left(3-\frac{\log (x)}{\log (3)}\right)=\log (3x)$

$3\log (3)-\log(x)=\log (3x)$

$3\log (3)-\log (x)-\log(3x)=0$

$\log (27)+\log (\frac{1}{x})+\log (\frac{1}{3x})=0$

$\log \left(\frac{27}{x(3x)}\right)=0$

$\log \left(\frac{9}{x^2}\right)=0$

$\frac{9}{x^2}=1$

$9=x^2$

$x=\pm 3$

$-3$ is rejected. So the only solution is $3$

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4 Answers

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Using another approach, recall that $\log_b x=\frac{\log_c x}{\log_cb}$.

Now, for $b=3x$ and $c=3$ we have

$$\log_{3x}3=\frac{\log_33}{\log_3 3x}=\frac{\log_33}{\log_3 3+\log_3x}=\frac{1}{1+\log_3x}$$

Let $y=\log_3 x$. Then, the equation

$$\left(3-\log_3x\right)\log_{3x}3=1\implies\frac{3-y}{1+y}=1\implies y=1\implies x=3$$

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After getting $$3\log(3)-\log(x)=\log(3x),$$ having $$\log\left(\frac{3^3}{x}\right)=\log(3x),$$ i.e. $$\frac{3^3}{x}=3x$$ might be easier.

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Hint:

note that: $$ log_{3x}3=\dfrac{\log_3 3}{\log_3 3x}=\dfrac{1}{1+\log_3 x} $$ so your equation become: $$ 3-\log_3x=1+\log_3x $$ that is easely solved as $\log_3 x=1$

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Here is another one: we must have $x>0, x\neq \frac{1}{3}$, so assume it.

$$\left(3-\log_3 x\right)\log_{3x} 3=\log_3 (27/x)\log_{3x} (3)$$

$$=\frac{\log_3(27/x)\overbrace{\log_3(3)}^{1}}{\log_3(3x)}=1$$

$$\iff \log_3(27/x)=\log_3(3x)\iff 27/x=3x$$

$$\iff x^2=9\iff x=3$$

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