Question

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

Answer 1

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 ,

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

Answer 2

$\purple{\tt{Solution :-}}$

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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 .

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

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

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

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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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