Logarithmic Equation solve log_(3x)3+log_(x/3)3=5/12

[Yankel]

Dear all,

I wish to solve the following logarithmic equation:

\[log_{3x}3+log_{\frac{x}{3}}3=\frac{5}{12}\]

My intuition was to start with changing the base of both logarithms to 10 (or any other number), but couldn't continue from there. Can you assist please ? Is there a meaning to the fact that both bases involves 3 in them ?

Thank you in advance.

 

[HOI]

Yes, you certainly want the two logarithms to the same base but there is nothing special about base 10. Since one logarithm is already to base "3x" I would be inclined to change the other to base 3x also. If [tex]y= log_{x/3}(3)[/tex] then [tex]3= (x/3)^y= x^y/3^y[/tex]. So [tex]x^y= 3^{y+1}[/tex] and then [tex](3x)^y= 3^yx^y= 3^{2y+ 1}[/tex]. Taking the logarithm, base 3x, of both sides, [tex]y= log_{3x}(3^{2y+1})= (2y+1)log_{3x}(3)[/tex]. Solving that for y, [tex](1- 2log_{3x}(3))y= log_{3x}(3)[/tex] so [tex]y= log_{x/3}(3)= \frac{log_{3x}(3)}{1- 2log_{3x}(3)}[/tex]

The original equation, [tex]log_{3x}(3)+ log_{x/3}(3)= \frac{5}{12}[/tex] becomes [tex]log_{3x}(3)\left(1+ \frac{log_{3x}(3)}{1-2log_{3x}(3)}\right)= \frac{5}{12}[/tex].

To simplify, let [tex]y= log_{3x}(3)[/tex] so the equation is [tex]y\left(1+ \frac{y}{1- 2y}\right)= \frac{5}{12}[/tex]. Solve that for y then solve [tex]log_{3x}(3)= y[/tex] for x.