Is this logarithmic true?

Question

$$x^{\log_{10}\left(\frac{y}{z}\right)}y^{\log_{10}\left(\frac{z}{x}\right)}z^{\log_{10}\left(\frac{x}{y}\right)}=1$$

for $x, y, z>0$.

Please help me I couldn't figure this out for the life of me.

Answer 1

Hint. Recall basic properties of the logarithm. Since $a^b=10^{b\cdot\log_{10} (a)}$ then we have that $$x^{\log_{10}\left(\frac{y}{z}\right)}=10^{(\log_{10}(y)-\log_{10}(z))\cdot\log_{10}(x)}$$ Manipulate the other terms in a similar way and you should obtain that the final exponent of $10$ is zero.

Answer 2

You have

$$x^{\log_{10}(y/z)}=x^{\log_{10}(y)-\log_{10}(z)}=\frac{x^{\log_{10}(y)}}{x^{\log_{10}(z)}}.$$

So you get

$$x^{\log_{10}(y/z)}y^{\log_{10}(z/x)}z^{\log_{10}(x/y)}=\frac{x^{\log_{10}(y)}y^{\log_{10}(z)}z^{\log_{10}(x)}}{x^{\log_{10}(z)}y^{\log_{10}(x)}z^{\log_{10}(y)}}=1.$$

Answer 3

$x^ {(\log_{10}y- \log_{10}z)} * y^ {(\log_{10}z - \log_{10}x)} * z ^{( \log_{10}x - \log_{10}y)} = 1$

Introducing $\log$ to each term, \begin{align*} \implies & \log[x^ {(\log_{10}y- \log_{10}z)} * y^ {(\log_{10}z - \log_{10}x)} * z ^{( \log_{10}x - \log_{10}y)}] = \log_{10}1\\ \\ \implies& \log_{10}x^ {(\log_{10}y- \log_{10}z)} + \log_{10}y^ {(\log_{10}z - \log_{10}x)} + \log_{10}z ^{( \log_{10}x - \log_{10}y)} = \log_{10}1\\ \\ \implies& ( \log_{10}y- \log_{10}z) \log_{10}x + (\log_{10}z - \log_{10}x) \log_{10}y + (\log_{10}x - \log_{10}y) \log_{10}z = 0\\ \\ \implies&\log_{10}x \cdot \log_{10}y - \log_{10}x \cdot \log_{10}z + \log_{10}y \cdot \log_{10}z - \log_{10}x \cdot \log_{10}y +\log_{10}x \cdot\log_{10}z \\ &- \log_{10}y \cdot \log_{10}z = 0\\ \\ \implies & 0 = 0\end{align*}

Answer 4

$$ \begin{align} \log_{10}\left(x^{\log_{10}\left(\frac yz\right)}y^{\log_{10}\left(\frac zx\right)}z^{\log_{10}\left(\frac xy\right)}\right) &=\log_{10}(x)\left(\log_{10}(y)-\log_{10}(z)\right)\\ &+\log_{10}(y)\left(\log_{10}(z)-\log_{10}(x)\right)\\ &+\log_{10}(z)\left(\log_{10}(x)-\log_{10}(y)\right)\\[3pt] &=0 \end{align} $$