Simplify the expression: $a^{\log {\sqrt \frac bc}}×b^{\log {\sqrt \frac ca}}×c^{\log {\sqrt \frac ab}}$

Question

My problem is

Simplify the expression:$$a^{\log {\sqrt \frac bc}}×b^{\log {\sqrt \frac ca}}×c^{\log {\sqrt \frac ab}}$$

Here $a,b,c \in \mathbb {R^+}$

My way:

$$\begin{cases} \frac bc=e^x \\ \frac ca=e^y \\ \frac ab=e^z \\ \end{cases} \Rightarrow e^{x+y+z}=1 \Rightarrow x+y+z=0 $$

$$a^{\log {\sqrt \frac bc}}×b^{\log {\sqrt \frac ca}}×c^{\log {\sqrt \frac ab}} = \sqrt {a^{\log e^x}} ×\sqrt {b^{\log e^y}}×\sqrt {c^{\log e^z}}=\sqrt {a^x×b^y×c^z}=\sqrt {a^x×\frac{a^y}{e^{yz}}×a^ze^{yz}}=\sqrt {a^{x+y+z}}=\sqrt {a^0}=1$$

Is this method correct and is there a better way?

Thank you!

Answer 1

Technically correct.

But we can try as follows:

As $\log \dfrac bc=\log b-\log c$ when all the logarithm remain defined.

$$a^{\log\sqrt{\dfrac bc}}=a^{\dfrac{\log b-\log c}2}=\left(e^{\log a}\right)^{\dfrac{\log b-\log c}2}$$

$$=e^{\dfrac{\log a\log b-\log c\log a}2}$$

Answer 2

It's correct.

I like the following writing. $$\prod_{cyc}a^{\ln\sqrt{\frac{b}{c}}}=e^{\frac{1}{2}\sum\limits_{cyc}\ln{a}(\ln{b}-\ln{c})}=e^0=1.$$

Answer 3

$$\begin{align}M&=a^{\log {\sqrt \frac bc}}×b^{\log {\sqrt \frac ca}}×c^{\log {\sqrt \frac ab}}\\ \log M&=\frac 12\big(\log b-\log c\big)\log a+\frac 12 \big(\log c-\log a\big)\log b+\frac 12 \big(\log b-\log c\big)\log a\\&=0\\ M&=a^{\log {\sqrt \frac bc}}×b^{\log {\sqrt \frac ca}}×c^{\log {\sqrt \frac ab}}=\color{red}1\end{align}$$ This works for any base.