Question

Eliminate x and y by coefficient method::
8^x × 4^y = 32
81^x ÷ 27^y = 3​

Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

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$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{a^m\times{a^n}\:=\:a^{m\:+\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\purple{\dfrac{a^m}{a^n}\:=\:a^{m\:-\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\orange{\dfrac{1}{x^n}\:=\:x^{-n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\color{peru}{(a^m)^n\:=\:a^{m\times{n}}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\red{ {x}^{0} = 1}}}}} \\ \end{gathered}$

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$\huge \orange{AηsωeR} ✍$

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☆ Let us consider the first equation :

$\bf \:  {8}^{x} \times {4}^{y} = 32$

$\sf \:  ⟼ {2}^{3x} \times {2}^{2y} = {2}^{5}$

$\sf \:  ⟼ {2}^{3x + 2y} = {2}^{5}$

☆ On comparing, we get

$\bf \:  ⟼ 3x + 2y = 5 \: - - - (1)$

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☆ Let us consider the second equation :

$\bf \:  {81}^{x} \div {27}^{y} = 3$

$\sf \:  ⟼ {3}^{4x} \div {3}^{3y} = 3$

$\sf \:  ⟼ {3}^{4x - 3y} = {3}^{1}$

☆ On comparing, we get

$\bf\implies \:4x - 3y = 1 \: - - - (2)$

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☆ Solving (1) and (2) equation to get x and y using method of eliminations

☆Multiply (1) by 3 and (2) by 2, we get

$\bf \:  ⟼ 9x + 6y = 15 \: - - - (3) \\ \bf \:  ⟼ 8x - 6y = 2 \: \: \: - - - (4)$

☆On adding equation (3) and equation (4), we get

$\bf\implies \:17x = 17$

$\bf\implies \:x = 1 \: - - - - (5)$

☆ On substituting x = 1, in equation (2), we get

$\bf \:  ⟼ 4 \times 1 - 3y = 1$

$\bf \:  ⟼ 4 - 3y = 1$

$\bf \:  ⟼ - 3y = - 3$

$\bf\implies \:y = 1 \: - - - - (6)$

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$\large \boxed{\bf \: Hence, \: the \: solution \: is \: x = 1 \: and \: y = 1}$

Answer 2

$\LARGE{\bf{\underline{\underline{Solution:-}}}}$

${\sf{\underline{{We \ are \ given \ two \ equations:}}}}$

• $\sf{{8^{x}\times 4^{y} = 32}}$

• $\sf{{81^{x}\times 27^{y} = 3}}$

• $\sf{{8^{x}\times 4^{y} = 32}}$

$\implies \sf{{ (2^{3})^{x}. (2^{2})^{y} =2^{5}}}$

$\implies \sf{{2^{3x}.2^{2y} = 2^{5}}}$

$\implies \sf{{3x + 2y = 5}}$$\sf{{----- (1)}}$

• $\sf{{81^{x}\times 27^{y} = 3}}$

$\implies \sf{{ (3^{4})^{x}\div (3^{3})^{y} = 3}}$

$\implies \sf{{ 3^{4x}\div 3^{3y} = 3^{1}}}$

$\implies \sf{{4x - 3y = 1}}$$\sf{{----(2)}}$

${\sf{\underline{{Now \ let \ us \ eliminate \ y \ by \ using \ coefficient \ method:}}}}$

$\sf{{12x + 8y = 20}}$

$\sf{{12x - 9y = 3}}$

$\sf{{-}}$

$\sf{{--------}}$

$\sf{{17y = 17}}$

$\implies \sf{{y = 1}}$

$\sf{{--------}}$

${\sf{{{As \ we \ know \ y \ = \ 1, \ so \ x:}}}}$

$\sf{{3x + 2y = 5}}$

$\implies \sf{{3x + 2(1) = 5}}$

$\implies \sf{{3x + 2 = 5}}$

$\implies \sf{{3x = 5 - 2}}$

$\implies \sf{{3x = 3}}$

$\implies \sf{{x = 1}}$

${\sf{\underline{{Verification:-}}}}$

$\sf{{8^{x}\times 4^{y} = 32}}$

$\implies \sf{{8^{1}\times 4^{1} = 32}}$

$\implies \sf{{8\times 4 = 32}}$

$\implies \sf{{32 = 32}}$

$\therefore \sf{{L.H.S=R.H.S}}$

$\LARGE{\bf{\underline{\underline{Required \ Answer :-}}}}$

• $\sf{{x = 1}}$

• $\sf{{y = 1}}$