7
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ Also $x=y=-\sqrt{3}$ and $x=-\sqrt{\frac{3}{2}},y=-\sqrt{6}$. $\endgroup$ – Joel Reyes Noche Jan 4 '13 at 15:18
  • 1
    $\begingroup$ @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
2
$\begingroup$

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 $$ x=t+s, \\ y=t-s. $$

$\endgroup$
  • $\begingroup$ Very good solution! Thank yo $\endgroup$ – naoufelabs Jan 4 '13 at 20:12
5
$\begingroup$

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 $f' <0$.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.