If a² + b² = 23ab show that,
log { ⅕ (a+b) } = ½ (log a + log b).
Answers
EXPLANATION.
⇒ a² + b² = 23ab [Given].
To prove.
⇒ ㏒(a + b)/5 = 1/2(㏒a + ㏒b).
Method = 1.
⇒ ㏒(a + b)/5 = 1/2(㏒a + ㏒b).
As we know that,
We can write equation as,
⇒ 2㏒(a + b)/5 = ㏒(a) + ㏒(b).
As we know that,
Formula of :
⇒ ㏒ₐMN = ㏒ₐM + ㏒ₐN.
⇒ ㏒ₐN^(α) = α㏒ₐN (α any real number).
Using this formula in the equation, we get.
⇒ 2㏒(a + b)/5 = ㏒(ab).
⇒ [(a + b)/5]² = ab.
⇒ (a² + b² + 2ab)/25 = ab.
⇒ a² + b² + 2ab = 25ab.
⇒ a² + b² = 25ab - 2ab.
⇒ a² + b² = 23ab.
Hence proved.
Method = 2.
⇒ a² + b² = 23ab [Given].
To prove.
⇒ ㏒(a + b)/5 = 1/2(㏒a + ㏒b).
As we know that,
Formula of :
⇒ (a + b)² = a² + b² + 2ab.
Using this formula in the equation, we get.
⇒ (a + b)² = a² + b² + 2ab.
Put the value of (a² + b² = 23ab) in the equation, we get.
⇒ (a + b)² = 23ab + 2ab.
⇒ (a + b)² = 25ab.
Taking log on both sides of the equation, we get.
⇒ ㏒[(a + b)²] = ㏒(25ab).
As we know that,
Formula of :
⇒ ㏒ₐMN = ㏒ₐM + ㏒ₐN.
⇒ ㏒ₐN^(α) = α㏒ₐN (α any real number).
Using this formula in the equation, we get.
⇒ 2㏒(a + b) = ㏒(25) + ㏒(a) + ㏒(b).
⇒ 2㏒(a + b) = ㏒(5)² + ㏒(a) + ㏒(b).
⇒ 2㏒(a + b) = 2㏒(5) + ㏒(a) + ㏒(b).
⇒ 2㏒(a + b) - 2㏒(5) = ㏒(a) + ㏒(b).
As we know that,
Formula of :
⇒ ㏒ₐM/N = ㏒ₐM - ㏒ₐN.
Using this formula in the equation, we get.
⇒ 2[㏒(a + b) - ㏒(5)] = ㏒(a) + ㏒(b).
⇒ 2[㏒(a + b)/5] = ㏒(a) + ㏒(b).
⇒ ㏒(a + b)/5 = 1/2[㏒(a) + ㏒(b)].
Hence Proved.