Question

xraised to power p =y raised to power q = xy raised to power pq. prove that p+q=1​

Answer 1

Given,

$x^p=y^q=(xy)^{pq}$

We have to prove: $p+q=1.$

Let,

$x^p=y^q=(xy)^{pq}=k$

$x^p=k\quad\implies\quad p=\log_xk=\dfrac {\log k}{\log x}\\\\\\y^q=k\quad\implies\quad q=\log_yk=\dfrac {\log k}{\log y}\\\\\\\begin {aligned}&(xy)^{pq}=k\\\\\\\implies\ &pq=\log_{xy}k\\\\\\&pq=\dfrac {\log k}{\log (xy)}\\\\\\&pq=\dfrac {\log k}{\log x+\log y}\\\\\\&\dfrac {1}{pq}=\dfrac {\log x+\log y}{\log k}\\\\\\&\dfrac {1}{pq}=\dfrac {\log x}{\log k}+\dfrac {\log y}{\log k}\\\\\\&\dfrac {1}{pq}=\dfrac {1}{p}+\dfrac {1}{q}\\\\\\&\dfrac {1}{pq}=\dfrac {p+q}{pq}\\\\\\\implies\ &p+q=1\end {aligned}$

Hence Proved!