I have a system of equations $xy=3$ and $4^{x^2}+2^{y^2}=72$ whose solution I know is $x=y=\sqrt 3$, but what are the steps to solve it?

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    $\begingroup$ The expressions 4^x^2 and 2^y^2 are ambiguous. Do you mean $4^{(x^2)}$ and $2^{(y^2)}$, or do you mean $(4^x)^2$ and $(2^y)^2$? Perhaps you meant $4x^2$ and $2y^2$? $\endgroup$ – Zev Chonoles Jan 4 '13 at 11:54
  • $\begingroup$ I mean: 4^(x^2) and 2^(y^2). $\endgroup$ – naoufelabs Jan 4 '13 at 13:21
  • $\begingroup$ There are other solutions, such as $x=\sqrt{\frac{3}{2}},y=\sqrt{6}$. $\endgroup$ – Ofir Jan 4 '13 at 15:11
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    $\begingroup$ Also $x=y=-\sqrt{3}$ and $x=-\sqrt{\frac{3}{2}},y=-\sqrt{6}$. $\endgroup$ – Joel Reyes Noche Jan 4 '13 at 15:18
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    $\begingroup$ @naoufelabs - Is the answer you accepts meets your request? As far as I see, it solves a special case. $\endgroup$ – Ofir Jan 5 '13 at 0:46

The first attempt to try to solve a system of two (or more variables) is naturally assumed that a possible solution is one in which all variables are equal.

Then set $x=y=t$ and solve de equation \begin{cases} xy=t\cdot t=3\\ 4^{x^2}+2^{y^2}=\big(2^{t^2}\big)^2+2^{t^2}=72 \end{cases}

By $t^2=3$ we have $t=\pm\sqrt{3}$. Substituting into the another equation we have that two solutions are $(x,y)=(+\sqrt{3},+\sqrt{3})$ and $(x,y)=(-\sqrt{3},-\sqrt{3})$. Or by $\big(2^{t^2}\big)^2+2^{t^2}=72$ whe have $2^{t^2}=\frac{-1+\sqrt{289}}{2}=8$.

To further investigate other solutions you might assume $$ x=t+s, \\ y=t-s. $$

  • $\begingroup$ Very good solution! Thank yo $\endgroup$ – naoufelabs Jan 4 '13 at 20:12

The only solutions are $\pm(\sqrt{3},\sqrt{3})$ and $\pm(\sqrt{1.5},\sqrt{6})$. Proof:

Substitute $a=2x^2, b=y^2$. This becomes $2^a + 2^b = 72$. The relation between $a,b$ becomes $ab=2(xy)^2=18$ and $a,b>0$.

Let $f(x)=2^{x} + 2^{18/x}$. We are interested in positive solutions to $f(x)=72$. Since $f(x)=f(\frac{18}{x})$, we can restrict ourselves to $x \le \sqrt{18}$.

One solution is $f(3)=2^3 + 2^6=72$, which corresponds to $2x^2=3,y^2=6$, i.e. $\pm(\sqrt{1.5},\sqrt{6})$.

Another solution is recovered - $6=\frac{18}{3}$, which corresponds to $2x^2=6,y^2=3$, i.e. $\pm(\sqrt{3},\sqrt{3})$.

I'll show that $f(x)=72$ has no solutions for $0<x<\sqrt{18}$ other than $x=3$. The proof will be by showing that $f$ is decreasing in that interval:

$$f'(x)=\ln 2 ( 2^x -\frac{18}{x^2} 2^{\frac{18}{x}})$$

For $0<x<\sqrt{18}$, $$1<\frac{18}{x^2}, 2^{x} < 2^{\frac{18}{x}}$$ So $2^x < \frac{18}{x^2}2^{\frac{18}{x}}$, proving $f' <0$.

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    $\begingroup$ Very good your solution! $\endgroup$ – MathOverview Jan 4 '13 at 15:50
  • $\begingroup$ It can be generalized to equations of the form $a^{cx^r}+b^{dy^r}=n$, where $a > 1,b >1,c,d,n$ are fixed and $xy$ has a fixed value. $\endgroup$ – Ofir Jan 4 '13 at 15:55
  • $\begingroup$ Very good solution! Thank you $\endgroup$ – naoufelabs Jan 4 '13 at 20:11

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