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There are two real functions $f$ and $g$ for $ x, y> 0, a >1$ such that $f(x) = a^x + a^{-x}$ and $g(x) = a^x - a^{-x}$.

Find the $x$ satisfying $f(x)f(y)= 8$ and $g(x)g(y)=4$.

I've tried this by putting the $X$ and $Y$ as instead of the $ a^x$ and $a^y$ respectively. But the calculatuion process really complicated. Is there any efficient way?

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Thanks.

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  • $\begingroup$ Show us what you did. $\endgroup$ – Wuestenfux Aug 14 at 11:57
  • $\begingroup$ How is $y$ defined? $\endgroup$ – Parcly Taxel Aug 14 at 11:59
  • $\begingroup$ I eidted my post @Wuestenfux $\endgroup$ – se-hyuck yang Aug 14 at 12:03
  • $\begingroup$ @ParclyTaxel the $y$ is also positive real number like the $x $. I edited my question. $\endgroup$ – se-hyuck yang Aug 14 at 12:03
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    $\begingroup$ You might notice that your equation $X^2+Y^2=2XY$ has a consequence which may help a lot. $\endgroup$ – Mark Bennet Aug 14 at 12:04
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You will get the system $$a^{x+y}+\frac{1}{a^{x+y}}+a^{x-y}+\frac{1}{a^{x-y}}=8$$ and $$a^{x+y}+\frac{1}{a^{x+y}}-a^{x-y}-\frac{1}{a^{x-y}}=4$$ Now substitute. For intstance $$u+\frac{1}{u}+v+\frac{1}{v}=8$$ $$u+\frac{1}{u}-\left(v+\frac{1}{v}\right)=4$$

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Speacial thanks for the Mark bennet.

enter image description here

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Notice that

$$f(x)=2\cosh(x\log a),\\g(x)=2\sinh(x\log a).$$

Then

$$f(x)f(y)-g(x)g(y)=4\cosh((x-y)\log a)=4$$

so that $$x=y.$$

Also

$$f(x)f(y)+g(x)g(y)=4\cosh((x+y)\log a)=12$$ so that

$$x=y=\frac{\text{arcosh }3}{2\log a}=\log_a\sqrt{3+2\sqrt2}=\log_a(1+\sqrt2).$$

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