If-log-2x-1-18-log-18-1-3y-log-3y-1-2x-find-3x-2y-

Question Number 100675 by bobhans last updated on 28/Jun/20

If \log_{2 x} (\frac{1}{18}) = \log_{18} (\frac{1}{3 y}) = \log_{3 y} (\frac{1}{2 x})

find 3 x - 2 y

Commented by bramlex last updated on 28/Jun/20

\Leftrightarrow 2 x = 3 y = 18 \to {\begin{cases} x = 9 \\ y = 6 \end{cases}

∴ 3 x - 2 y = 27 - 12 = 15

Answered by 1549442205 last updated on 28/Jun/20

\log_{2 x} (\frac{1}{18}) = \log_{2 x} (1) - \log_{2 x} 18 = - \log_{2 x} 18

= \log_{3 y} (\frac{1}{2 x}) = \log_{3 y} 1 - \log_{3 y} (2 x) = - \log_{3 y} (2 x)

\log_{18} (\frac{1}{3 y}) = \log_{18} 1 - \log_{18} (3 y) = - \log_{18} (3 y)

, so from the hypothesis we get :

\log_{2 x} 18 = \log_{3 y} (2 x) = \log_{18} (3 y) = a . So

{\begin{cases} {(2 x)}^{a} = 18 (1) \\ {(3 y)}^{a} = 2 x (2) (*) \\ 18^{a} = 3 y (3) \end{cases}

From (2) we get {(2 x)}^{a} = {[{(3 y)}^{a}]}^{a} = {(3 y)}^{a^{2}} (4)

From (3) we get {(3 y)}^{a^{2}} = {(18^{a})}^{a^{2}} = 18^{a^{3}} (5)

From (4), (5) we get {(2 x)}^{a} = 18^{a^{3}} (6)

From (1) and (6) we obtain

18 = 18^{a^{3}} \Rightarrow a^{3} = 1 \Leftrightarrow a = 1 . Replace into (*)

we get {\begin{cases} 3 y = 2 x \\ 2 x = 18 \\ 18 = 3 y \end{cases} \Leftrightarrow {\begin{cases} x = 9 \\ y = 6 \end{cases}

Therefore, 3 x - 2 y = 3 \times 9 - 2 \times 6 = 15