$x$, $y$, $z$ and $p$ are four literals. The literal $p$ is used as a base of three multiplying functions, the reciprocal of subtraction of any two of remaining three literals with sequence form becomes an exponent to each base of the each multiplying exponential function. Similarly, another reciprocal of subtraction of any two from three becomes the whole power of each exponential term. It is also followed a sequence.

The value of entire multiplying function is required to find in this problem.

${\Bigg(\large p^{\normalsize \dfrac{1}{x-y}}\Bigg)}^{\dfrac{1}{x-z}}$ $\times$ ${\Bigg(\large p^{\normalsize \dfrac{1}{y-z}}\Bigg)}^{\dfrac{1}{y-x}}$ $\times$ ${\Bigg(\large p^{\normalsize \dfrac{1}{z-x}}\Bigg)}^{\dfrac{1}{z-y}}$

01

Each exponential function that contains a whole power can be simplified by applying the power rule of exponent of an exponential function formula.

$=\,\,$ $\large p^{\normalsize \Bigg[\dfrac{1}{x-y} \times \dfrac{1}{x-z}\Bigg]}$ $\times$ $\large p^{\normalsize \Bigg[\dfrac{1}{y-z} \times \dfrac{1}{y-x}\Bigg]}$ $\times$ $\large p^{\normalsize \Bigg[\dfrac{1}{z-x} \times \dfrac{1}{z-y}}\Bigg]$

$=\,\,$ $\large p^{\normalsize \Bigg[\dfrac{1}{(x-y)(x-z)}\Bigg]}$ $\times$ $\large p^{\normalsize \Bigg[\dfrac{1}{(y-z)(y-x)}\Bigg]}$ $\times$ $\large p^{\normalsize \Bigg[\dfrac{1}{(z-x)(z-y)}\Bigg]}$

02

The three multiplying exponential functions are having same base. So, they can be merged as a single multiplying exponential function by using product law of exponents.

$=\,\,$ $\large p^{\normalsize \dfrac{1}{(x-y)(x-z)} \large \,+\, \normalsize \dfrac{1}{(y-z)(y-x)} \large \,+\, \normalsize \dfrac{1}{(z-x)(z-y)}}$

$=\,\,$ $\large p^{\normalsize \dfrac{1}{(x-y)[-(z-x)]} \large \,+\, \normalsize \dfrac{1}{(y-z)[-(x-y)]} \large \,+\, \normalsize \dfrac{1}{(z-x)[-(y-z)]}}$

$=\,\,$ $\large p^{\normalsize \dfrac{-1}{(x-y)(z-x)} \large \,+\, \normalsize \dfrac{-1}{(y-z)(x-y)} \large \,+\, \normalsize \dfrac{-1}{(z-x)(y-z)}}$

$=\,\,$ $\large p^{\normalsize \dfrac{-y+z-z+x-x+y}{(x-y)(y-z)(z-x)}}$

$=\,\,$ $\require{cancel} \large p^{\normalsize \dfrac{-\cancel{y}+\cancel{z}-\cancel{z}+\cancel{x}-\cancel{x}+\cancel{y}}{(x-y)(y-z)(z-x)}}$

$=\,\,$ $\large p^{\normalsize \dfrac{0}{(x-y)(y-z)(z-x)}}$

03

According to zero power rule, the value of any exponential function that contains $0$ as power, is equal to $1$.

$=\,\,$ $\large p^{\normalsize 0}$

$=\,\,$ $1$

Therefore, the value of entire function is one and it is the required solution for this exponent problem in mathematics.

Latest Math Topics

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved