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Mama L [17]
3 years ago
5

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

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
4 0

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 <em>b</em> > 0 and <em>b</em> ≠ 1. I picked the natural base, <em>e</em>.

Then

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

as required.

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