 QUESTION

# Simplify the given logarithmic expression:${x^{\log y - \log z}}.{y^{\log z - \log x}}.{z^{\log x - \log y}}$.

Hint: Here, we will simplify the given expression by using the properties of logarithms like $\log (a.b) = \log a + \log b$ and $\log {a^b} = b\log a$ to find the value of x.
Consider the given expression as X and take $\log$both sides,
$X = {x^{\log y - \log z}}.{y^{\log z - \log x}}.{z^{\log x - \log y}} \\ \Rightarrow \log X = \log ({x^{\log y - \log z}}.{y^{\log z - \log x}}.{z^{\log x - \log y}}) \\$
Here we can use $\log (a.b) = \log a + \log b$
$\log X = \log ({x^{\log y - \log z}}.{y^{\log z - \log x}}.{z^{\log x - \log y}}) \\ \Rightarrow \log X = \log ({x^{\log y - \log z}}) + \log ({y^{\log z - \log x}}) + \log ({z^{\log x - \log y}}) \\ \Rightarrow \log X = (\log y - \log z)\log (x) + (\log z - \log x)\log (y) + (\log x - \log y)\log (z){\text{ }}[\because \log {a^b} = b\log a] \\ \Rightarrow \log X = \log y\log x - \log z\log x + \log z\log y - \log x\log y + \log x\log z - \log y\log z \\ \Rightarrow \log X = 0 \\ \Rightarrow X = {10^0}{\text{ }}[{\text{taking anti - log}}] \\ \Rightarrow X = 1 \\$