Question

Solve for x : log(x-1) + log(x+1) = log 1 a base 2

Answer 1

Using property,

$log(x) + log(y) = log(xy)$

LHS=

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

and RHS =
$log_{2}(1) = log(1)$

Equating LHS and RHS,
${x}^{2} - 1 = 1 \\ {x}^{2} = 2 \\ x = + \sqrt{2}$

Ignoring the negative value since the argument takes only positive ones.

Answer 2

Given :-

$log(x - 1) + log(x + 1) = log_{2}(1)$

To Find :-

Value of x.

$\rule{200}{1}$

LHS :-

$= log(x - 1) + log(x + 1)$

We know the Property -

• $log(x) + log(y) = log(xy)$

So ,

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

RHS :-

$= log_{2}(1) \\ = log(1)$

Now equating the LHS and RHS -

$log( {x}^{2} - 1 ) = log(1) \\ {x}^{2} - 1 = 1 \\ {x}^{2} = 1 + 1 \\ {x}^{2} = 2 \\ \boxed{ x = \sqrt{2} }$

$\rule{200}{1}$

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