Question

$log(x-1)+log(x+1) = log8 then x = how much$
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Answer 1

Solution:-

$\rm \: log(x - 1) + log(x + 1) = log8$

By using this identity

$\rm \: log(x) + log(y) = log(xy)$

We get

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

$\rm \ log(x - 1) (x + 1) = log(8)$

log of lhs and rhs side get cancel out

$\rm \: (x + 1)(x - 1) = 8$

Using this identity

$\rm \: (a - b)(a + b) = ( {a}^{2} - {b}^{2} )$

We get

$\rm(x {}^{2} - 1) = 8$

$\rm \: {x}^{2} = 8 + 1$

$\rm \: {x}^{2} = 9$

$\rm \: x = \sqrt{9}$

$\rm \: x = \pm3$

Answer 2

$\sf{\large{\bigstar{\bold{\pink{\boxed{\boxed{Solution}}}}}}}$

$\sf\implies log(x-1)+log(x+1) = log8$

$\sf{\red{\boxed{\bold{log\: a + log\: b = log \:ab }}}}$

Therefore :-

$\sf\implies log(x-1)+log(x+1) = log ( x - 1 )( x + 1 )$-----( i )

$\sf\implies log(x-1)+log(x+1) = log8$-------( ii )

• In eq i and ii LHS is equal therefore LHS will also be equal

$\sf\implies log(x-1)(x+1) = log8$

• Cancelling log on either side

$\sf\implies ( x - 1 )( x + 1 ) = 8$

$\sf{\red{\boxed{\bold{( a + b ) ( a - b ) = a^2 - b^2}}}}$

$\sf\implies x^2 - 1 = 8$

$\sf\implies x^2 = 8 + 1$

$\sf\implies x^2 = 9$

$\sf\implies x = \sqrt 9$

$\sf\leadsto \dag x = \pm 3$

$\sf{\green{\boxed{\bold{\dag x = \pm 3 }}}}$