Question

10. If 3^x+y= 81 and 81^x-y=3^8, then the values of x and y are respectively​

Answer 1

Step-by-step explanation:

${3}^{x + y} = 81 \\ = > {3}^{x + y} = {3}^{4} \\ = > x + y = 4.......(i) \\ \\ {81}^{x - y} = {3}^{8} \\ = > {3}^{4(x - y)} = {3}^{8} \\ = > 4(x - y) = 8 \\ = > x - y = 2.....(ii) \\ \\ from \: (i) \: and \: (ii) \\ x + y = 4 \\ x - y = 2 \\ ................. \\ 2x = 6 \\ = > x = 3 \: put \: in \: (i) \\ 3 + y = 4 = > y = 1$

Answer 2

Step-by-step explanation:

$\mathsf{Given : 3^{x + y} = 81}$

$\mathsf{\bigstar\;\;81\;can\;be\;written\;as : 3^4}$

$\mathsf{\implies 3^{x + y} = 3^4}$

$\bigstar\;\;\textsf{When Bases are same on both sides, Exponents should be equal}$

$\mathsf{\implies x + y = 4\;-----\;(1)}$

$\mathsf{Given : 81^{x - y} = 3^{8}}$

$\mathsf{\implies (3^4)^{x - y} = 3^{8}}$

$\mathsf{\implies 3^{4(x - y)} = 3^{8}}$

$\mathsf{\implies 4(x - y) = 8}$

$\mathsf{\implies (x - y) = \dfrac{8}{4}}$

$\mathsf{\implies x - y = 2\;-----\;(2)}$

$\textsf{Adding Equations (1) and (2), We get :}$

$\mathsf{\implies (x + y) + (x - y) = 4 + 2}$

$\mathsf{\implies x + x + y - y = 6}$

$\mathsf{\implies 2x = 6}$

$\mathsf{\implies x = 3}$

$\textsf{Substituting x = 3 in Equation (1), We get :}$

$\mathsf{\implies 3 + y = 4}$

$\mathsf{\implies y = 4 - 3}$

$\mathsf{\implies y = 1}$

Answers :

★  x = 3

★  y = 1