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[tex] {( \sqrt{3} )}^{x + y} = 9 \\ {( \sqrt{2} )}^{x – y} = 32[/tex] find the value of 2x + y.
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[tex] {( \sqrt{3} )}^{x + y} = 9 \\ {( \sqrt{2} )}^{x – y} = 32[/tex]
find the value of 2x + y.
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Mathematics
1 year
2021-09-03T15:08:25+00:00
2021-09-03T15:08:25+00:00 1 Answers
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[tex] \bf \purple{ \underline{Given :-}}[/tex]
[tex] • \: {( \sqrt{3} )}^{x + y} = 9\: \: \: \: (i)[/tex]
[tex]• \: {( \sqrt{2} )}^{x – y} = 32 \: \: \: \: (ii)[/tex]
[tex] \\ [/tex]
[tex] \bf \purple{ \underline{To \: Find :- }}[/tex]
[tex] • {\sf{The \: value \: of}} \: \: 2x+ y.[/tex]
[tex] \\ [/tex]
[tex] \huge\bf \purple{ \underline{Solution :- }}[/tex]
[tex] \sf{From \: equation \: (i), }[/tex]
[tex] {( \sqrt{3} )}^{x + y} = 9 [/tex]
[tex]⇒ {( \sqrt{3} )}^{x + y} = ({ \sqrt{3} })^{4} [/tex]
[tex] ⇒ x + y = 4 \: \: \: \: (iii)[/tex]
[tex] \\ [/tex]
[tex] \sf{From \: the \: equation \: (ii)}[/tex]
[tex]{( \sqrt{2} )}^{x – y} = 32[/tex]
[tex] ⇒( { \sqrt{2} })^{x – y} =( { \sqrt{2} })^{10} [/tex]
[tex]⇒x – y = 10 \: \: \: \: (iv)[/tex]
[tex] \\ [/tex]
[tex]\sf{We \: have \: to \: add \: equation \: (iii) \: and \: equation \: (iv)}[/tex]
[tex] x + y + x – y = 4 + 10[/tex]
[tex] ⇒2x = 14[/tex]
[tex] ⇒x = 7[/tex]
[tex] \\ [/tex]
[tex] \sf{Subtracting \: equation \: (iii) \: from \: equation \: (ii), }[/tex]
[tex]x – y – x – y = 10 – 4[/tex]
[tex]⇒ – 2y = 6[/tex]
[tex]⇒y = – 3[/tex]
[tex] \\ [/tex]
[tex] \bf\therefore x = 7 \: \: and \: \: y = – 3[/tex]
[tex] \\ [/tex]
[tex] { \sf{Th e \: value \: of }}= 2x +y [/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =2 \times 7 + ( – 3)[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =14 – 3[/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =11[/tex]
[tex] \bf \red{Hence, \: the \: value \: of \: 2x + y \: is \: 11. }[/tex]