Question

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

Answer 1

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.

Complete step-by-step answer:
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 \\\$$

Hence, the value of the given expression is 1.

Note: If the base of the logarithm is not given, we can consider the base as 10 i.e. common logarithm. These are the circular questions. In this question we are playing with x, y, z. We can add one-two more variables and form a bigger question. Nevertheless, the concept to solve the problem will not change.