Question

If 4^x+y= 1 and 4^x-y=4, find x and y

Answer 1

Appropriate Question :- Find x and y

$\sf \: {4}^{x + y} = 1 \: \: \: and \: \: \: {4}^{x - y} = 4$

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

Given that,

$\sf \: {4}^{x + y} = 1$

$\sf \: {4}^{x + y} = {4}^{0}$

$\bf\implies \:x + y = 0 - - - - (1)$

Further given that

$\sf \: {4}^{x - y} = 4$

$\sf \: {4}^{x - y} = {4}^{1}$

$\bf\implies \:x - y = 1 - - - (2)$

On adding equation (1) and (2), we get

$\sf \: 2x = 1$

$\bf\implies \:x = \dfrac{1}{2}$

On substituting the value of x in equation (1), we get

$\sf \: \dfrac{1}{2} + y = 0$

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

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

Additional Information :-

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