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  • 15-09-2021
  • Mathematics
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AlgebraicAardvark
AlgebraicAardvark AlgebraicAardvark
  • 15-09-2021

If [tex]3x-y=12[/tex], then [tex]y=3x-12[/tex].  Substituting that into [tex]\dfrac{8^x}{2^y}[/tex] would give us...

         [tex]\begin{aligned}\dfrac{8^x}{2^y} &= \dfrac{8^x}{2^{3x-12}} \\[0.9em]&= \dfrac{8^x}{2^{3x}\cdot2^{-12}}\\[0.9em]&= \dfrac{8^x}{(2^3)^x\cdot2^{-12}}\\[0.9em]&= \dfrac{8^x}{8^{x}\cdot2^{-12}}\\[0.9em]&= \dfrac{1}{1\cdot2^{-12}}\\[0.9em]&= \dfrac{1}{2^{-12}}\\[0.9em]&= 2^{12}\end{aligned}[/tex]

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