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Question-181923




Question Number 181923 by manxsol last updated on 02/Dec/22
Answered by SEKRET last updated on 02/Dec/22
   xlog2(y)=6         y=2^(6/x)    (((x+y)/(x−y)))log2(y)=4    (6/x)= ((4x−4y)/(x+y))      6x+6y=4x^2 −4xy    y(6+4x)=4x^2 −6x     y= ((x(2x−3))/((2x+3)))
$$\:\:\:\boldsymbol{\mathrm{xlog}}\mathrm{2}\left(\boldsymbol{\mathrm{y}}\right)=\mathrm{6}\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}=\mathrm{2}^{\frac{\mathrm{6}}{\boldsymbol{\mathrm{x}}}} \\ $$$$\:\left(\frac{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}}\right)\boldsymbol{\mathrm{log}}\mathrm{2}\left(\boldsymbol{\mathrm{y}}\right)=\mathrm{4} \\ $$$$\:\:\frac{\mathrm{6}}{\boldsymbol{\mathrm{x}}}=\:\frac{\mathrm{4}\boldsymbol{\mathrm{x}}−\mathrm{4}\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}}\:\:\:\:\:\:\mathrm{6}\boldsymbol{\mathrm{x}}+\mathrm{6}\boldsymbol{\mathrm{y}}=\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{xy}} \\ $$$$\:\:\boldsymbol{\mathrm{y}}\left(\mathrm{6}+\mathrm{4}\boldsymbol{\mathrm{x}}\right)=\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{\mathrm{x}}\:\:\:\:\:\boldsymbol{\mathrm{y}}=\:\frac{\boldsymbol{\mathrm{x}}\left(\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3}\right)}{\left(\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{3}\right)} \\ $$
Commented by Frix last updated on 02/Dec/22
Yes.
$$\mathrm{Yes}. \\ $$

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