Question

If a²+b²=7ab,prove that log
{1/3(a+b)}=1/2(loga+logb)

Answer 1

Step-by-step explanation:

a^2 + b^2 = 7a*b

(a+b)^2- 2a*b = 7a*b { a^2 + b^2 = (a+b)^ -2ab}

(a+b)^2 = 9a*b.

(a+b) = 3√ab. ....(1). { square root both sides}

to prove - log{1/3(a+b)} = 1/2( loga + logb)

proof - taking LHS

log{1/3(a+b)} = log{1/3(3√ab)}

= log√ab

taking RHS

1/2(loga + logb) = 1/2(logab) { logm + logn = log mn}

= log√ab { logm^n = nlogn}

LHS = RHS

Hence proved....