Logarithm Equality.

Question

$$\sqrt {\log_a(ax)^{\frac{1}{4}} + \log_x(ax)^{\frac{1}{4}}} + \sqrt {\log _a{(\frac{x}{a})^{\frac{1}{4}}} + \log_x (\frac{a}{x})^\frac{1}{4}} = a,$$ 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.

Answer 1

$$\log_a(ax)^{\frac14}+\log_x(ax)^{\frac14}=\frac{\log_aa+\log_ax+\log_xa+\log_xx}4$$

$$=\frac14\left(2+\frac{\log a}{\log x}+\frac{\log x}{\log a}\right)$$

$$=\frac14\left(\sqrt{\frac{\log a}{\log x}}+\sqrt{\frac{\log x}{\log a}}\right)^2$$

$$=\frac14\left(\frac1{\sqrt{\log_ax}}+\sqrt{\log_a x}\right)^2$$

$$\implies\sqrt{\log_a(ax)^{\frac14}+\log_x(ax)^{\frac14}}=\frac12\left(\frac1{\sqrt{\log_ax}}+\sqrt{\log_a x}\right) $$

and similarly, $$\sqrt{\log_a(x/a)^{\frac14}+\log_x(a/x)^{\frac14}}=\frac12\left|\frac1{\sqrt{\log_ax}}-\sqrt{\log_a x}\right|$$

Answer 2

$$\sqrt { \log_a\left((ax)^\frac 14\right) + \log_x\left((ax)^\frac 14\right) } + \sqrt { \log_a\left(\left(\frac xa\right)^\frac 14\right) + \log_x\left(\left(\frac ax \right)^\frac 14\right) } = a$$

$$\sqrt { \frac 14\log_a\left(ax\right) + \frac 14\log_x\left(ax\right) } + \sqrt { \frac 14\log_a\left(\frac xa\right) + \frac 14\log_x\left(\frac ax\right) } = a$$

$$\sqrt { \log_a\left(ax\right) + \log_x\left(ax\right) } + \sqrt { \log_a\left(\frac xa\right) + \log_x\left(\frac ax\right) } = 2a$$

$$\sqrt { \log_a(a) + \log_a(x) + \log_x(a) + \log_x(x) } + \sqrt { \log_a(x) - \log_a(a) + \log_x(a) - \log_x(x) } = 2a$$

$$\sqrt { 1 + \underbrace{\log_a(x) + \log_x(a)}_z + 1 } + \sqrt { \log_a(x) - 1 + \log_x(a) - 1 } = 2a$$

$$\sqrt { z + 2 } + \sqrt { z - 2 } = 2a$$

$$z + 2 + 2\sqrt {z^2 - 4} + z - 2 = 4a^2$$

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