Question

If log(a+b)/3=log (ab)/2,
prove that:
a²+b²=7ab

Answer 1

a² + b² = 7ab  

=> a² + b² + 2ab = 7ab + 2ab  

=> (a + b)² = 9ab  

=> (a + b) = 3√(ab)  

=> (a + b)/3 = √ab  

=> log{(a + b)/3} = log (√ab)  

=> log{(a + b)/3} = (1/2)log(ab)  

=> log{(a + b)/3} = (1/2)(loga+ logb)  

=> log{(a + b)/3} = (loga)/2+ (logb)/2  

=> log{(a + b)/3} - (loga)/2- (logb)/2 = 0

=> log{(a + b)/3} - [log ab/2] = 0

=> log{(a + b)/3} = log(ab)/2