MHB 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 y= log_{x/3}(3) then 3= (x/3)^y= x^y/3^y. So x^y= 3^{y+1} and then (3x)^y= 3^yx^y= 3^{2y+ 1}. Taking the logarithm, base 3x, of both sides, y= log_{3x}(3^{2y+1})= (2y+1)log_{3x}(3). Solving that for y, (1- 2log_{3x}(3))y= log_{3x}(3) so y= log_{x/3}(3)= \frac{log_{3x}(3)}{1- 2log_{3x}(3)}

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

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