Math, asked by tishaarora1302, 1 year ago

If 2^(x)=10^(y)=50^(z), show that y=2xz/z+x

Answers

Answered by Saby123
13

In the above Question , the following information is given -

 \sf{ 2^x = { 10 }^ y = { 50 }^ z }

To find -

Show that -

 \sf{ y = \dfrac{ 2xz}{ x + z } }

Solution -

Here ,

 \sf{ 2^x = { 10 }^ y = { 50 }^ z }

Let this be equal to k .

So ,

 \sf{ 2^x = { 10 }^ y = { 50 }^ z = k  }

Now , let us express 2, 10 and 50 in terms of k .

 \sf{ 2^x = k } \\ \\ \sf{ \implies { 2 = k^ { \frac{1}{x} } }}

 \sf{ { 10 } ^ y = k } \\ \\ \sf{ \implies { 10 = k^ { \frac{1}{y} } }}

 \sf{ { 50 } ^ z = k } \\ \\ \sf{ \implies { 50 = k^ { \frac{1}{z} } }}

5 = [ 10 / 2 ]

Substituting the values of k ,

 \sf{ 5 = k ^ { \frac{ 1 }{ y } - \frac{1}{x} }}

Now ,

10 × 5 = 50

Substituting the values of k ,

 \sf{ k ^ { \frac{ 1 }{ y } - \frac{1}{x} } \times k^ { \frac{1}{y} } = k^ { \frac{1}{z} } } \\ \\ \sf{\implies {  k ^ { \frac{ 2 }{ y } - \frac{1}{x} } = k^ { \frac{1}{z} } }}  \\ \\ \sf{ \implies { \dfrac{ 2 }{ y } - \dfrac{ 1}{ x} = \dfrac{ 1}{ z } }} \\ \\ \sf{ \implies { \dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z} }} \\ \\ \sf{ \implies { \dfrac{ 2}{y} = \dfrac{ x + z }{ xz } }} \\ \\ \sf{ \implies { y = \dfrac{2xz}{ x + z }} } \\ \\ \sf{ \bold { Hence \: Shown \: }}

Answered by Anonymous
42

\purple{\tt{Solution :-}}

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Let \sf{ 2^x = { 10 }^ y = { 50 }^ z } be equal to k .

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Expressing 2, 10 and 50 in terms of k .

\begin{lgathered}\tt{ 2^x = k } \\ ↠\tt{ 2 = k^ { \frac{1}{x} } }\end{lgathered}

\begin{lgathered} ↠\tt{ { 10 } ^ y = k } \\ ↠ \tt{ 10 = k^ { \frac{1}{y} } }\end{lgathered}

\begin{lgathered}↠\tt{ { 50 } ^ z = k } \\  ↠\tt { 50 = k^ { \frac{1}{z}  }}\end{lgathered}

5 = \tt{\frac{10}{2}}\\

Substituting the values of k ,

\tt{ 5 = k ^ { \frac{ 1 }{ y } - \frac{1}{x} }}

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Then,

10 × 5 = 50

Substituting the values of k ,

\tt{ k ^ { \frac{ 1 }{ y } - \frac{1}{x} } \times k^ { \frac{1}{y} } = k^ { \frac{1}{z} } } \\ ↠\tt { k ^ { \frac{ 2 }{ y } - \frac{1}{x} } = k^ { \frac{1}{z} } } \\  ↠\tt{ \dfrac{ 2 }{ y } - \dfrac{ 1}{ x} = \dfrac{ 1}{ z } } \\ \tt↠ { \dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z} } \\ ↠\tt { \dfrac{ 2}{y} = \dfrac{ x + z }{ xz } } \\ ↠\red{\tt{ y = \dfrac{2xz}{ x + z }}}

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