Math, asked by madhav5245, 1 month ago

Solve the equation for x

 \frac{1 -  2{(log {x}^{2} )}^{2} }{logx - 2 {(logx)}^{2} }  = 1 \\

Answers

Answered by mathdude500
23

\large\underline{\sf{Solution-}}

Given equation is

\sf \: \dfrac{1 - 2{(log {x}^{2} )}^{2} }{logx - 2 {(logx)}^{2} } = 1 \\  \\

can be further rewritten as

\sf \: \dfrac{1 - 2{(2 \: logx )}^{2} }{logx - 2 {(logx)}^{2} } = 1 \\  \\

Let assume that

\sf \: logx = y \\  \\

So, above expression can be rewritten as

\sf \: \dfrac{1 - 2 {(2y)}^{2} }{y - 2 {y}^{2} }  = 1 \\  \\

\sf \: \dfrac{1 - 2 \times  {4y}^{2}  }{y - 2 {y}^{2} }  = 1 \\  \\

\sf \: \dfrac{1 - {8y}^{2}  }{y - 2 {y}^{2} }  = 1 \\  \\

\sf \: 1 -  {8y}^{2} = y - 2 {y}^{2}  \\  \\

\sf \:  {6y}^{2} + y - 1 = 0 \\  \\

\sf \:  {6y}^{2} + 3y - 2y - 1 = 0 \\  \\

\sf \: 3y(2y + 1) - 1(2y + 1) = 0 \\  \\

\sf \: (2y + 1)(3y - 1) = 0 \\  \\

\bf\implies \:y =  - \dfrac{1}{2} \:  \: or  \: \: \dfrac{1}{3}  \\  \\

Now,

\sf \: logx =  - \dfrac{1}{2}  \: \bf\implies \:x =  {\bigg( 10\bigg) }^{ -  \: \dfrac{1}{2} }  \\  \\

and

\sf \: logx = \dfrac{1}{3}  \: \bf\implies \:x =  {\bigg( 10\bigg) }^{\: \dfrac{1}{3} }  \\  \\

\rule{190pt}{2pt}

 {{ \mathfrak{Additional\:Information}}}

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x)  = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} )  = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} )  = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b)  = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx}  = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx}  =  {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \\  \end{array} }}\end{gathered}\end{gathered}\end{gathered}

Answered by ot7xbangtonboyz
31

Given equation is:-

  \sf \frac{1 - 2{(log {x}^{2} )}^{2} }{logx - 2 {(logx)}^{2} } = 1 \\

can we for the rewritten as

 \sf \frac{1 - 2{(log {x}^{2} )}^{2} }{logx - 2 {(logx)}^{2} } = 1 \\

Let assume that,

 \sf \:  logx = y

so above expression can be written as

  \sf  \frac{1 - 2( {2)}^{2} }{y -  {2y}^{2}  } = 1 \\

 \sf \:  \frac{1 - 2 \times  {4y}^{2} }{y -  {2y}^{2} }  = 1 \\

 \sf \:  \frac{1 -  {8y}^{2} }{y -  {2y}^{2} }  = 1 \\

 \implies \sf \: 1 -  {8y}^{2}  = y -  {2y}^{2}

 \sf \implies \ {6y}^{2}  + y - 1 = 0

 \: \sf \implies \:  {6y}^{2}  + 3y - 2y - 1 = 0

 \ \sf \implies \: 3y(2y + 1) - 1(2y + 1) = 0

 \rm \implies \: (2y + 1)(3y - 1) = 0

 \rm \implies \: y =  -  \frac{1}{2}  \: or \:  \frac{1}{3}  \\

Now,

 \sf \:  logx =  - 1 \longmapsto \: x = (10){}^ { -  \frac{1}{2}  }  \\

and

 \sf \:  logx =  - 1 \longmapsto \: x = (10){}^ { -  \frac{1}{3}  }  \\

 \rule{333pt}{5.0pt}

 \begin{array}{ |c| } \hline \\ \text{ \: more \: formulae} \\  \hline \text{ \:  logx(x) = 1}  \\  \hline  \ \: logx( {x}^{y} )= y \\  \hline {e}^{ylogx}  =  {x}^{y}  \\ \hline log1 = 0 \\  \hline\end{array}

 \sf{ \purple{ \pmb{ \frak{ \: be \: brainly}}}}

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