Math, asked by romibhamani786, 9 months ago

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

Answers

Answered by Anonymous
36

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

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