# Solving $f(x)f(y)= 8$ and $g(x)g(y)=4$, where $f(x) = a^x + a^{-x}$ and $g(x) = a^x - a^{-x}$

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?

Thanks.

• Show us what you did. – Wuestenfux Aug 14 at 11:57
• How is $y$ defined? – Parcly Taxel Aug 14 at 11:59
• I eidted my post @Wuestenfux – se-hyuck yang Aug 14 at 12:03
• @ParclyTaxel the $y$ is also positive real number like the $x$. I edited my question. – se-hyuck yang Aug 14 at 12:03
• You might notice that your equation $X^2+Y^2=2XY$ has a consequence which may help a lot. – Mark Bennet Aug 14 at 12:04

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 $$u+\frac{1}{u}+v+\frac{1}{v}=8$$ $$u+\frac{1}{u}-\left(v+\frac{1}{v}\right)=4$$

Speacial thanks for the Mark bennet.

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

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