Question

prove that (logx²-logx) × log(1/x) + (logx)²=0

$prove \: that \: ( log{x}^{2} - logx) \times log( \frac{1}{x} ) + { (logx) }^{2} = 0$
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Answer 1

To prove :

(logx² - logx)•log(1/x) + (logx)² = 0

Proof :

=> LHS = (logx² - logx)•log(1/x) + (logx)²

=> LHS = (2logx - logx)•log(1/x) + (logx)²

=> LHS = logx•log(1/x) + (logx)²

=> LHS = logx•[ log(1/x) + logx ]

=> LHS = logx•[ log{x^(-1)} + logx ]

=> LHS = logx•[ -logx + logx ]

=> LHS = logx•0

=> LHS = 0

=> LHS = RHS

Hence proved .

Properties used :

• logA + logB = logAB

• logA - logB = log(A/B)

• logAⁿ = n·logA

• log(1/A) = -logA

Answer 2

Step-by-step explanation:

(log n)^a -- log an

log n^a -- a log n

log a + log b -- log ( a.b )