Logarithm Equality. $$\sqrt {\log_a(ax)^{\frac{1}{4}} + \log_x(ax)^{\frac{1}{4}}}  + \sqrt {\log _a{(\frac{x}{a})^{\frac{1}{4}}} + \log_x (\frac{a}{x})^\frac{1}{4}}  = a,$$ for $a>0$ and different than 1... I keep getting $a = 1$, but that cannot be. I use log identities to transform the above into $$\sqrt {\frac{1}{2} + \frac{2\ln | ax |}{4\ln | ax |}}  + \sqrt {\frac{1}{4}\frac{2\ln | ax |}{\ln | ax |} - \frac{1}{2}}  = a $$ which means $a = 1$. Maybe I am overlooking something, but I do not see what. 

$$\sqrt {
    \log_a\left((ax)^\frac 14\right)
  + \log_x\left((ax)^\frac 14\right)
} + \sqrt {
  \log_a\left(\left(\frac xa\right)^\frac 14\right)
  + \log_x\left(\left(\frac ax \right)^\frac 14\right)
}  = a$$
$$\sqrt {
    \frac 14\log_a\left(ax\right)
  + \frac 14\log_x\left(ax\right)
} + \sqrt {
    \frac 14\log_a\left(\frac xa\right)
  + \frac 14\log_x\left(\frac ax\right)
}  = a$$
$$\sqrt {
    \log_a\left(ax\right)
  + \log_x\left(ax\right)
} + \sqrt {
    \log_a\left(\frac xa\right)
  + \log_x\left(\frac ax\right)
}  = 2a$$
$$\sqrt {
    \log_a(a) + \log_a(x)
  + \log_x(a) + \log_x(x)
} + \sqrt {
    \log_a(x) - \log_a(a)
  + \log_x(a) - \log_x(x)
}  = 2a$$
$$\sqrt {
    1 + \underbrace{\log_a(x) + \log_x(a)}_z + 1
} + \sqrt {
    \log_a(x) - 1 + \log_x(a) - 1
}  = 2a$$
$$\sqrt {
    z + 2
} + \sqrt {
    z - 2
}  = 2a$$
$$z + 2 + 2\sqrt {z^2 - 4} + z - 2 = 4a^2$$
$$2\sqrt {z^2 - 4}= 4a^2 - 2z$$
$$4z^2 - 16 = 16a^4 - 16a^2z^2 + 4z^2$$
$$\frac{a^4 + 1}{a^2} = z^2$$
$$\frac{a^4 + 1}{a^2} = \left(\frac{\log(a)}{\log(x)} + \frac{\log(x)}{\log(a)}\right)^2$$
$$\frac{a^4 + 1}{a^2} = \frac{\log(a)^2}{\log(x)^2} + 2 + \frac{\log(x)^2}{\log(a)^2}$$
$$\frac{a^4 -2a^2 + 1}{a^2} = \frac{\log(a)^2}{\underbrace{\log(x)^2}_y} + \frac{\log(x)^2}{\underbrace{\log(a)^2}_w}$$
$$yw\left(\frac{a^2 - 1}{a}\right)^2 = w^2 + y^2$$
$$\text{...Etc}$$
