Question

Solve for x

$\boxed{\sf log(x-1)+log(x+1)=log_21}$


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Answer 1

Answer:

Value of x is √2

Step-by-step explanation:

Given :- Log(x-1) + log (x+1) = log₂1

Using property of logarithm

Using property of logarithm log ₐx + logₐy = logₐ (xy)

• ∴ log{(x-1)(x +1) = log₂1

• log(x²-1) = log₂1 = 0

{∵ property logₐ1 = 0}

we know that if base isn't given

( we take base as 10 )

• Since, Log₁₀(x²-1) = 0

using property logₐx = y means x = aʸ

• x²-1 = 10⁰

• x² -1 = 1

• x² = 2

• x = ±√2

But, We know log cannot be have negative value

so value of x should be √2

Answer 2

Answer:

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Ans=log(x−1)+log(x+1)= log

2

1

⇒log(x−1)+log(x+1)=0

⇒log[(x−1)(x+1)]=0

⇒(x−1)(x+1)=1....(Since log1=0) ⇒x

2

−1=1

⇒x

2

=2

⇒x=±

2

−

2

cannot be possible , since log of a negative number is not defined.

So, x=

2