[tex]\huge{\textbf{\textsf{{☆ QUE}}{\purple{ST}}{\pink{ION ☆} \: {{}{:}}}}}[/tex]
If 3x – y = 12,
What is the value of
[tex] \frac{ {8}^{x} }{{2}^{y} } [/tex]
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Answer:
Answers for x and y according to the first equation, 3x-y=12, are:
(3,-3), (4,0), (5,3) etc.
Plug them into the second equation, f(x,y)=(8^x)/(2^y):
f(3,-3) = (8^3)/(2^-3) = (8^3)/(1/8) = 8^4
f(4,0) = (8^4)/(2^0) = (8^4)/1 = 8^4
f(5,3) = (8^5)/(2^3) = (8^5)/8 = 8^4
The value is 8^4 but this can be simplified
8 = 2^3
(2^3)^4 = 2^12
= 4096
Given : [tex]\qquad \sf \leadsto \:\: 3x – y = 12 \:\: \qquad [/tex]
Exigency To Find : The value of [tex]\dfrac{8^x}{2^y} [/tex] .
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
❍ Let’s 3x – y = 12 as Equation. 1 .
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Finding Value of [tex]\bf \dfrac{8^x}{2^y} [/tex] :
[tex]\qquad \dashrightarrow \:\: \sf \dfrac{8^x}{2^y} \:\:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\: \sf \dfrac{8^x}{2^y} \:\:\\\\[/tex]
[tex]\dag\:\:\sf{ As,\:We\:know\:that\::}\\[/tex]
[tex]\qquad \dag\:\:\bigg\lgroup \sf{ 2^3 \:\: or\:\:2 \times 2\times 2 \: =\:\:8 }\bigg\rgroup \\\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Applying \: this \::}}\\[/tex]
[tex]\qquad \dashrightarrow \:\: \sf \dfrac{8^x}{2^y} \:\:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\: \sf \dfrac{(2^3)^x}{2^y} \:\:\\\\[/tex]
[tex]\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\bf \:\: Law’s\:of\:Exponent\:\:: \\ [/tex]
[tex]\qquad \dag\:\:\bigg\lgroup \sf{ (a^m)^n = a \:^{m \times n} }\bigg\rgroup \\\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Applying \: this \::}}\\[/tex]
[tex]\qquad \dashrightarrow \:\: \sf \dfrac{(2^3)^x}{2^y} \:\:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\: \sf \dfrac{(2)^{3 \times x}}{2^y} \:\:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\: \sf \dfrac{(2)^{3x}}{2^y} \:\:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\: \sf \dfrac{2^{3x}}{2^y} \:\:\\\\[/tex]
[tex]\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\bf \:\: Law’s\:of\:Exponent\:\:: \\ [/tex]
[tex]\qquad \dag\:\:\bigg\lgroup \sf{ \dfrac{a^m}{a^n} = a \:^{m – n} }\bigg\rgroup \\\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Applying \: this \::}}\\[/tex]
[tex]\qquad \dashrightarrow \:\: \sf \dfrac{2^{3x}}{2^y} \:\:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\: \sf 2^{3x- y} \:\:\\\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Equation \:\:1 \: Value\::}}\\[/tex]
[tex]\qquad \dag\:\:\bigg\lgroup \sf{ Equation \:\:1 \:=\:3x – y = 12 }\bigg\rgroup \\\\[/tex]
[tex]\qquad \dashrightarrow \:\: \sf 2^{12} \:\:\\\\[/tex]
[tex]\qquad \dashrightarrow \pmb{\underline{\purple{\:2^{12} }} }\:\:\bigstar \\[/tex]
⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \sf \:Hence \:The \:value \:of\:\dfrac{8^x }{2^y}\:is\:\bf 2^{12}}}\\[/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
[tex]\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}\\\\[/tex]
[tex]\qquad \qquad \boxed{\begin{array}{cc}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m – n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{array}}[/tex]