Question

log(x-1)+log(x+1)=log of 1 at base 2

Answer 1

Answer:

Step-by-step explanation:

log(x+1) + log (x-1)  = $log_{2}$1

⇒ log [(x+1)(x-1)] = 0

⇒ log (x²-1) = log 1

⇒ x²-1 = 1

⇒ x² = 2

⇒ x = √2

Answer 2

Here is Your answer..❤

$log(x + 1) \: + log \: (x - 1)$

$log \: (x + 1)(x - 1) = 0$

$log \: {x}^{2} = log(1)$

${x}^{2} - 1 = 1$

$x = \sqrt{2}$

Hope it helps...

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