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If (2^(x^(2)))/(2^(y^(2)))=(1)/(8) and x+y=-3, what is the value of x-y ?

If 2z22y2=18 \frac{2^{z^{2}}}{2^{y^{2}}}=\frac{1}{8} and x+y=−3 x+y=-3 , what is the value of x−y x-y ?

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Full solution

Q. If 2z22y2=18 \frac{2^{z^{2}}}{2^{y^{2}}}=\frac{1}{8} and x+y=−3 x+y=-3 , what is the value of x−y x-y ?
  1. Rewrite equation: Rewrite the equation 2x22y2=18\frac{2^{x^2}}{2^{y^2}} = \frac{1}{8} as 2x2−y2=182^{x^2 - y^2} = \frac{1}{8}.
  2. Recognize base: Recognize that 18\frac{1}{8} is 2−32^{-3}, so we have 2x2−y2=2−3.2^{x^{2} - y^{2}} = 2^{-3}.
  3. Set exponents equal: Since the bases are the same, set the exponents equal to each other: x2−y2=−3x^{2} - y^{2} = -3.
  4. Factor left side: Factor the left side as (x+y)(x−y)=−3(x+y)(x-y) = -3.
  5. Substitute x+yx+y: Substitute x+y=−3x+y=-3 into the equation: (−3)(x−y)=−3(-3)(x-y) = -3.
  6. Solve for x−yx-y: Divide both sides by −3-3 to solve for x−yx-y: x−y=1x-y = 1.

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