Question

The sum of solutions of the equation log2x.log4x.log6x =
log2x.log4x + log4x.log6x + logex.log2x is equal to​

Answer 1

We have to find the sum of solutions of log2(x) × log4(x) × log6(x) = log2(x) × log4(x) + log4(x) × log6(x) + log6(x) × log2(x)

Solution : log2(x) × log4(x) × log6(x) = log2(x) × log4(x) + log4(x) × log6(x) + log6(x) × log2(x)

⇒logx/log2 × logx/log4 × logx/log6 = logx/log2 × logx/log4 + logx/log4 × logx/log6 + logx/log6 × logx/log2

⇒(logx)³/(log2 log4 log6) = (logx)²/(log2 log4) + (logx)²/(log4 log6) + (logx)²/(log6 log2)

⇒(logx)³/(log2 log4 log6) = (logx)²[1/(log2 log4) + 1/(log4 log6) + 1/(log6 log2)]

⇒(logx)³/(log2 log4 log6) = (logx)²[(log2 + log4 + log6)/(log2 log4 log6)]

⇒(logx)³ = (logx)²[log(2 × 4 × 6)]

⇒(logx)²[logx - log48] = 0

⇒logx = 0, log48

x = 1 or 48

Therefore the sum of solutions of given equation is = 1 + 48 = 49