Question

If p = q^r, q=r^p and r=p^q, prove that prq=1.​

Answer 1

Rewrite each given equation as

$p=q^r\implies \log_p(p)=\log_p(q^r)\implies 1=r\log_p(q)\implies r=\dfrac1{\log_p(q)}=\dfrac{\ln(q)}{\ln(p)}$

$q=r^p\implies \log_q(q)=\log_q(r^p)\implies 1=p\log_q(r)\implies p=\dfrac1{\log_q(r)}=\dfrac{\ln(r)}{\ln(q)}$

$r=p^q\implies \log_r(r)=\log_r(p^q)\implies 1=q\log_r(p)\implies q=\dfrac1{\log_r(p)}=\dfrac{\ln(p)}{\ln(r)}$

where each of the last equalities follows from the change-of-base identity,

$\log_m(n)=\dfrac{\log_b(n)}{\log_b(m)}$

for any base b > 0 and b ≠ 1. I picked the natural base, e.

Then

$prq=\dfrac{\ln(r)}{\ln(q)}\times\dfrac{\ln(q)}{\ln(p)}\times\dfrac{\ln(p)}{\ln(r)}=1$

as required.