(2)^(1/(log _x (((243)/x)))) = ((log


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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
	$get 5−2log_x 3=u { ((log_x (((243)/x))=5log_x 3−1=((−5)/2)u+((23)/2) )),((log_x ((x^5 /9))=5−2log_x 3=u)) :} ⇒2^((1/2)(11.5−2.5u)) =u^(1/3) ⇒ get 2^(1.5) =e^(0.4A) ⇒e^(A(4.6−u)) =u⇒e^(4.6A) =ue^(Au) ⇒Ae^(4.6A) =Aue^(Au) Au=w(Ae^(4.6A) )⇒5−2log_x 3=(1/A)w(Ae^(4.6A) ) ⇒x=9^(((A/(5A−W)))) { ((A=((15)/4)ln2)),((W=w(Ae^(4.6A) ))) :}$
	$g e t 5 - 2 {l o g}_{x} 3 = u$ ${\begin{cases} {l o g}_{x} (\frac{243}{x}) = 5 {l o g}_{x} 3 - 1 = \frac{- 5}{2} u + \frac{23}{2} \\ {l o g}_{x} (\frac{x^{5}}{9}) = 5 - 2 {l o g}_{x} 3 = u \end{cases}$ $\Rightarrow 2^{\frac{1}{2} (11.5 - 2.5 u)} = u^{\frac{1}{3}} \Rightarrow g e t 2^{1.5} = e^{0.4 A}$ $\Rightarrow e^{A (4.6 - u)} = u \Rightarrow e^{4.6 A} = {u e}^{A u} \Rightarrow {A e}^{4.6 A} = {A u e}^{A u}$ $A u = w ({A e}^{4.6 A}) \Rightarrow 5 - 2 {l o g}_{x} 3 = \frac{1}{A} w ({A e}^{4.6 A})$ $\Rightarrow x = 9^{(\frac{A}{5 A - W})} {\begin{cases} A = \frac{15}{4} l n 2 \\ W = w ({A e}^{4.6 A}) \end{cases}$
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