Question

Given that p² + q² = 11qp where p and q are constants
show that
½(log p + log q) equals

a) and b) in the image​

Answer 1

Question:

Given that p² + q² = 11pq where p and q are constants. Show that

$\displaystyle{\sf\:\dfrac{1}{2}\:(\:\log\:p\:+\:\log\:q\:)\:equals}$

$\displaystyle{\sf\:a)\:\log\:\left(\:\dfrac{p\:-\:q}{3}\:\right)}$

$\displaystyle{\sf\:b)\:\log\:\left(\:\dfrac{p\:+\:q}{\sqrt{13}}\:\right)}$

Answer:

$\displaystyle{\sf\:a)\:{\boxed{\red{\sf\:\dfrac{1}{2}\:(\:\log\:p\:+\:\log\:q\:)\:=\:\log\:\left(\:\dfrac{p\:-\:q}{3}\:\right)\:}}}}$

$\displaystyle{\sf\:b)\:{\boxed{\blue{\sf\:\dfrac{1}{2}\:(\:\log\:p\:+\:\log\:q\:)\:=\:\log\:\left(\:\dfrac{p\:+\:q}{\sqrt{13}}\:\right)\:}}}}$

Step-by-step-explanation:

a)

We have given that,

$\displaystyle{\sf\:p^2\:+\:q^2\:=\:11\:pq}$

$\displaystyle{\implies\sf\:p^2\:+\:q^2\:=\:2\:pq\:+\:9\:pq}$

$\displaystyle{\implies\sf\:p^2\:+\:q^2\:-\:2\:pq\:=\:9\:pq}$

$\displaystyle{\implies\sf\:(\:p\:-\:q\:)^2\:=\:9\:pq}$

$\displaystyle{\implies\sf\:(\:p\:-\:q\:)\:=\:\sqrt{9\:pq}\:\qquad\dots\:[\:Taking\:square\:roots\:]}$

$\displaystyle{\implies\sf\:(\:p\:-\:q\:)\:=\:3\:\sqrt{pq}}$

$\displaystyle{\implies\sf\:\dfrac{p\:-\:q}{3}\:=\:\sqrt{pq}}$

$\displaystyle{\implies\sf\:\log\:\left(\:\dfrac{p\:-\:q}{3}\:\right)\:=\:\log\:(\:\sqrt{pq}\:)\:\qquad\dots[\:Taking\:log\:on\:both\:sides\:]}$

$\displaystyle{\implies\sf\:\log\:\left(\:\dfrac{p\:-\:q}{3}\:\right)\:=\:\log\:[\:(\:pq\:)^{\frac{1}{2}}\:]}$

$\displaystyle{\implies\sf\:\log\:\left(\:\dfrac{p\:-\:q}{3}\:\right)\:=\:\dfrac{1}{2}\:\log\:(\:pq\:)\:\qquad\dots\:[\:\log\:(\:a^k\:)\:=\:k\:\log\:a\:]}$

$\displaystyle{\implies\sf\:\log\:\left(\:\dfrac{p\:-\:q}{3}\:\right)\:=\:\dfrac{1}{2}\:(\:\log\:p\:+\:\log\:q\:)\:\qquad\dots[\:\log\:(\:ab\:)\:=\:\log\:a\:+\:\log\:b\:]}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{1}{2}\:(\:\log\:p\:+\:\log\:q\:)\:=\:\log\:\left(\:\dfrac{p\:-\:q}{3}\:\right)\:}}}}$

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b)

Now,

$\displaystyle{\sf\:p^2\:+\:q^2\:=\:11\:pq}$

$\displaystyle{\implies\sf\:(\:p\:+\:q\:)^2\:-\:2\:pq\:=\:11\:pq}$

$\displaystyle{\implies\sf\:(\:p\:+\:q\:)^2\:=\:11\:pq\:+\:2\:pq}$

$\displaystyle{\implies\sf\:(\:p\:+\:q\:)^2\:=\:13\:pq}$

$\displaystyle{\implies\sf\:(\:p\:+\:q\:)\:=\:\sqrt{13\:pq}\:\qquad\dots[\:Taking\:square\:roots\:]}$

$\displaystyle{\implies\sf\:(\:p\:+\:q\:)\:=\:\sqrt{13}\:\times\:\sqrt{pq}}$

$\displaystyle{\implies\sf\:\dfrac{p\:+\:q}{\sqrt{13}}\:=\:\sqrt{pq}}$

$\displaystyle{\implies\sf\:\log\:\left(\:\dfrac{p\:+\:q}{\sqrt{13}}\:\right)\:=\:\log\:(\:\sqrt{pq}\:)\:\qquad\dots[\:Taking\:log\:on\:both\:sides\:]}$

$\displaystyle{\implies\sf\:\log\:\left(\:\dfrac{p\:+\:q}{\sqrt{13}}\:\right)\:=\:\log\:[\:(\:pq\:)^{\frac{1}{2}}\:]}$

$\displaystyle{\implies\sf\:\log\:\left(\:\dfrac{p\:+\:q}{\sqrt{13}}\:\right)\:=\:\dfrac{1}{2}\:\log\:(\:pq\:)\:\qquad\dots\:[\:\log\:(\:a^k\:)\:=\:k\:\log\:a\:]}$

$\displaystyle{\implies\sf\:\log\:\left(\:\dfrac{p\:+\:q}{\sqrt{13}}\:\right)\:=\:\dfrac{1}{2}\:(\:\log\:p\:+\:\log\:q\:)\:\qquad\dots[\:\log\:(\:ab\:)\:=\:\log\:a\:+\:\log\:b\:]}$

$\displaystyle{\therefore\:\underline{\boxed{\blue{\sf\:\dfrac{1}{2}\:(\:\log\:p\:+\:\log\:q\:)\:=\:\log\:\left(\:\dfrac{p\:+\:q}{\sqrt{13}}\:\right)\:}}}}$