Find x, if $ \log _{15}\left(\frac{2}{9}\right)^{\:}=\log _3\left(x\right)=\log _5\left(1-x\right) $

Question

So how can I find the value of x, if:

$$ \log _{15}\left(\frac{2}{9}\right)^{\:}=\log _3\left(x\right)=\log _5\left(1-x\right) $$

I tried switching everything to base 15, but that didn't work out well.

Answer 1

Solve the following system disjointedly not modifying the side with $x_1, x_2$ $$ \begin{cases} \log_{15}\left(\frac{2}{9}\right) &= \log_3 (x_1) \\ \log_{15}\left(\frac{2}{9}\right) &= \log_5 (1- x_2) \end{cases} $$

then check when solutions $x_1, x_2$ are same.

Answer 2

If $\log _{b}({y}) =x$, Then $\log _{cb}({y}{c}^{x}) =x $.

In the case where $\log _{15}(\frac{2}{9}) =\large\log_{3}({x})$

We have $\log _{15}(\frac{2}{9}) =\large\log_{3}[\,(\frac{2}{9})(\frac{1}{5})^{log_{15}(\frac{2}{9})}] = \log_3 (x) $.

Therefore $\large x= [\,(\frac{2}{9})(\frac{1}{5})^{log_{15}(\frac{2}{9})}]$ $\approx 0.5432531069176$

In the case where $\log _{15}(\frac{2}{9}) =\large\log_{5}({1-x})$

We have $\log _{15}(\frac{2}{9}) =\large\log_{5}[\,(\frac{2}{9})(\frac{1}{3})^{log_{15}(\frac{2}{9})}] = \log_5 (1-x) $.

So $\large [\,(\frac{2}{9})(\frac{1}{3})^{log_{15}(\frac{2}{9})}]= (1-x)$

Therefore $\large x= 1- [\,(\frac{2}{9})(\frac{1}{3})^{log_{15}(\frac{2}{9})}]$ $\approx 0.5909416450776$

In the case where $\log_{3}(x) = \log_{5}(1-x),\, x\approx0.5658150883239$

There is no case where the three equations are simultaneously equal