Question

if p=q^r q=r^p and r=p^q prove that pqr=1

Answer 1

Step-by-step explanation:

Let:-

$p = {q}^{r}$be eqn.1

$q = {r}^{p}$be eqn.2

$r = {p}^{q}$be eqn.3

Susbstituting the value of p from eqn.1 to eqn.3 ,

$r = {( {q}^{r} )}^{q} \: = {q}^{qr}$

Substituting the value of q from eqn.2 to eqn.3 ,

$r = ( { {r}^{p} )}^{qr} = ( {r})^{pqr}$

$= > {r}^{1} = {r}^{pqr}$

As bases are same in both the sides of this eqn.

$= > pqr = 1$

Answer 2

We can easily prove the statement that if $p=q^{r}$, $q = r^{p}$ and $r=p^{q}$ , then pqr=1.

What is exponent?

We know that a number's exponent indicates how many times it should be multiplied.

How to solve it?

Given that

$p=q^{r}$

$q = r^{p}$

Now, $r=p^{q}$

i.e. $r=(q^{r})^{q}$

i.e. $r=(q^{qr})$

i.e. $r=(r^{p})^{rq}$

i.e. $r=(r^{pqr})$

i.e. $\frac{r}{r^{pqr}} =1$

i.e. $r^{1-pqr} =r^{0}$ [Since $a^{0} = 1$, a is any number]

i.e. 1 - pqr = 0

i.e. pqr = 1

We proved our statement that if $p=q^{r}$, $q = r^{p}$ and $r=p^{q}$, then pqr=1.

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