Question

please answer this question​

Answer 1

Question :-

$\rm \: {5}^{x} = {7}^{y} = 1225, \: then \: find \: \dfrac{xy}{x + y}$

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

Given expression

$\rm \: {5}^{x} = {7}^{y} = 1225$

Let assume that

$\rm \: {5}^{x} = {7}^{y} = 1225 = k$

So, it means

$\rm \: {5}^{x} = k \: \rm\implies \:5 = {\bigg(k\bigg) }^{\dfrac{1}{x} } - - - (1)$

$\rm \: {7}^{y} = k \: \rm\implies \:7 = {\bigg(k\bigg) }^{\dfrac{1}{y} } - - - (2)$

and

$\rm \: 1225 = k$

$\rm \: 5 \times 5 \times 7 \times 7 = k$

$\rm \: {5}^{2} \times {7}^{2} = k$

On substituting the value of 5 and 7, we get

$\rm \: {\bigg(k \bigg) }^{\dfrac{2}{x} } \times {\bigg(k\bigg) }^{\dfrac{2}{y} } = k$

We know,

$\boxed{ \rm{ \: {a}^{x} \times {a}^{y} = {a}^{x + y} \: }}$

So, using this identity, we get

$\rm \: {\bigg(k \bigg) }^{\dfrac{2}{x} + \dfrac{2}{y} } = {k}^{1}$

We know,

$\boxed{ \rm{ \: {a}^{x} = {a}^{y}\rm\implies \:x = y \: }}$

So, using this result, we get

$\rm \: \dfrac{2}{x} + \dfrac{2}{y} = 1$

$\rm \: \dfrac{2y + 2x}{xy} = 1$

$\rm \: \dfrac{2(y + x)}{xy} = 1$

$\color{green}\rm\implies \:\boxed{ \rm{ \:\dfrac{xy}{x + y} = 2 \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$