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.