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[tex]\:\:\:\:\:\:\:\:\:\:\therefore\bf{x=3\:and\:y=1}[/tex]
Solution
[tex]\:\:\:\bf{(2^{x+y},\:3^{x-y})=(16,9)}[/tex]
[tex]\sf{Here,}[/tex]
[tex]\:\:\:\star\tt{\:\:2^{x+y}=16\:and\:3^{x-y}=9}[/tex]
[tex]\sf{Solving,}[/tex]
[tex]\:\:\:\implies\tt{\:\:2^{x+y}=16}[/tex]
[tex]\:\:\:\implies\tt{\:\:2^{x+y}=2^4}[/tex]
[tex]\:\:\:\implies\tt{\:\:x+y=4}[/tex]
[tex]\:\:\:\bf{\:\:x=4-y-----(i)}[/tex]
[tex]\sf{Solving,}[/tex]
[tex]\:\:\:\implies\tt{\:\:3^{x-y}=9}[/tex]
[tex]\:\:\:\implies\tt{\:\:3^{x-y}=3^2}[/tex]
[tex]\:\:\:\bf{\:\:x-y=2-----(ii)}[/tex]
[tex]\sf{Putting\:the\:value\:of\:(i)\:in\:(ii),}[/tex]
[tex]\:\:\:\implies\tt{\:\:4-y-y=2}[/tex]
[tex]\:\:\:\implies\tt{\:\:4-2y=2}[/tex]
[tex]\:\:\:\implies\tt{\:\:-2y=2-4}[/tex]
[tex]\:\:\:\implies\tt{\:\:\cancel{-}2y=\cancel{-}2}[/tex]
[tex]\:\:\:\implies\tt{\:\:y=\frac{\cancel{2}}{\cancel{2}}}[/tex]
[tex]\:\:\:\bf{\:\:y = 1}[/tex]
[tex]\sf{Putting\:y = 1\:in\:(i),}[/tex]
[tex]\:\:\:\implies\tt{\:\:x=4-1}[/tex]
[tex]\:\:\:\bf{\:\:x = 3}[/tex]
[tex]\:\:[/tex]