Question

If a^2+b^2=79ab then prove that 2 log(a+b)=4 log3+loga+logb

Answer 1

Question:-

If a^2 + b^2 = 79ab then prove that 2log(a+b) = 4log3+loga+logb

Note:-

• (a+b)^2 = a^2 + b^2 + 2ab
• (a-b)^2 = a^2 + b^2 - 2ab
• log(abc) = loga + logb + logc
• log(a/b) = loga - logb
• log(a^m) = mloga

Solution:-

• Considering the given statement ,a^2 + b^2 = 79ab
• a^2 + b^2 = 79ab
• Adding 2ab on both sides
• a^2 + b^2 + 2ab = 79ab + 2ab
• (a+b)^2 = 81ab
• Applying log on both sides,
• log ( (a+b)^2 ) = log (81ab)
• 2log(a+b) = log(3^4 . a . b)
• 2log(a+b) = log(3^4) + loga + logb
• 2log(a+b) = 4log3 + loga + logb

Hence ,

If a^2 + b^2 = 79ab, then 2log(a+b) = 4log3+loga+logb