2-1-log-x-243-x-log-x-x-5-9-1-3-

Question Number 163061 by tounghoungko last updated on 03/Jan/22

\sqrt[\log_{x} (\frac{243}{x})]{2} = \sqrt[3]{\log_{x} (\frac{x^{5}}{9})}

Answered by mahdipoor last updated on 03/Jan/22

g e t 5 - 2 {l o g}_{x} 3 = u

{\begin{cases} {l o g}_{x} (\frac{243}{x}) = 5 {l o g}_{x} 3 - 1 = \frac{- 5}{2} u + \frac{23}{2} \\ {l o g}_{x} (\frac{x^{5}}{9}) = 5 - 2 {l o g}_{x} 3 = u \end{cases}

\Rightarrow 2^{\frac{1}{2} (11.5 - 2.5 u)} = u^{\frac{1}{3}} \Rightarrow g e t 2^{1.5} = e^{0.4 A}

\Rightarrow e^{A (4.6 - u)} = u \Rightarrow e^{4.6 A} = {u e}^{A u} \Rightarrow {A e}^{4.6 A} = {A u e}^{A u}

A u = w ({A e}^{4.6 A}) \Rightarrow 5 - 2 {l o g}_{x} 3 = \frac{1}{A} w ({A e}^{4.6 A})

\Rightarrow x = 9^{(\frac{A}{5 A - W})} {\begin{cases} A = \frac{15}{4} l n 2 \\ W = w ({A e}^{4.6 A}) \end{cases}