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Solving a piecewise defined function for real solutions:

  • $x+y=65$
  • $\log_{2}{x} + \log_{4}{y} = 3$

So far I've changed $\log_{4}{y}$ in terms of base $2$ and then plugged that into the second equation. However, I've ended up with something like $y = x^{\frac{2}{64}}$ which seems more complicated than helpful to use the first equation.

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  • $\begingroup$ "Piecewise" ??? $\endgroup$ – Yves Daoust Dec 7 '15 at 17:33
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Your first step is correct. Rewrite the second equation as:

$$\log_2 x+\frac{1}{2}\log_2 y=3$$ $$2\log_2 x+\log_2 y=6$$ $$log_2x^2+\log_2y=6$$ $$log_2\left(x^2y\right)=6$$ $$x^2y=64$$ $$y=64x^{-2}$$ Sub this back into the first equation: $$x+64x^{-2}=65$$ $$x^3-65x^2+64=0$$ $$(x-1)(x^2-64x-64)=0$$ So $x=1$ or $x=32\pm8\sqrt{17}$ however $x>0$ so we can exclude the negative solution; then we get: $$(x,y)=(1,64), (32+8\sqrt{17},33-8\sqrt{17})$$

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Take the base $4$ antilogarithm of the second equation

$$x^2y=64.$$

As $y=65-x$,

$$x^2(65-x)-64=0.$$

By inspection, $x=1$ is a solution. Simplifying by $x-1$,

$$-x^2+64x+64=0,$$which you can solve.

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