Question

If log(p+q/3)=1/2(log p + log q) prove that p² + q² =7pq

Answer 1

If log[(p+q)/3]=1/2(log p + log q) then p² + q² is equal to 7pq.

Step-by-step explanation:

Required logarithm identities:

1. log (ab) = log a + log b
2. log aᵇ = b log a

We are given  

log[(p+q)/3] = ½[log p + log q]

Let’s solve the R.H.S first :-

½[log p + log q]

using identity (1)

= ½ * [log (pq)]

Using identity (2)

= log [(pq)^(1/2)] ……. (i)

Now, we will equate L.H.S = R.H.S as given in the question i.e.,

log[(p+q)/3] = ½[log p + log q]

substituting the value of R.H.S from eq. (i)

⇒ log[(p+q)/3] = log [(pq)^(1/2)]

cancelling log from both sides

⇒ (p+q)/3 = √[pq]

⇒ p + q = 3 * √(pq)

squaring both sides

⇒ (p +q)² = [3 * √(pq)]²

⇒ p² + 2pq + q² = 9pq

⇒ p² + q² = 9pq – 2pq

⇒ p² + q² = 7pq

Hence proved  

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