What is x if log2 (x^2+x+1)=logsquare root2 (x+1)?

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We have to solve log (2) (x^2+x+1) = log (sqrt 2)(x+1)

If y = log (2) (x^2+x+1) = log (sqrt 2)(x+1)

=> 2^y = x^2 + x + 1 and (sqrt 2)^y = x + 1

taking the square of both the sides of (sqrt 2)^y = x + 1,...

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We have to solve log (2) (x^2+x+1) = log (sqrt 2)(x+1)

If y = log (2) (x^2+x+1) = log (sqrt 2)(x+1)

=> 2^y = x^2 + x + 1 and (sqrt 2)^y = x + 1

taking the square of both the sides of (sqrt 2)^y = x + 1, we get

2^y = (x +1)^2 = x^2 + 2x + 1

So we have x^2 + x + 1 = x^2 + 1 + 2x

=> x = 0

Therefore x = 0.

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