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, 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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To find x if log2 (x^2+x+1)=log square root2 (x+1).
Solution:
log2 (x^2+x+1) = log (sqrt2) (x+1).
We know that log a^(1/2) x = {log(a) x}/log(a) a^(1/2) = 2log(a) x.
Therefore log (2) (x^2+x+1) = 2 log 2(x+1).
So log (2) x^2+x+x+1 = log(2) (x+1)^2.
Taking anti log we get:
x^2+x+1 = (x+1)^2 = x^2+2x+1.
x^2+x+1= x^2+2x+1.
=> 0 = x.
Therefore x= 0.
We'll solve the equation pretending that the bases of logarithms are 2 and sqrt2.
Though, before solving the equation, we'll impose the constraints of existence of logarithms.
x^2 + x + 1 > 0 for any real value of x.
x + 1 > 0
x > -1
Now, we'll solve the equation, changing the base sqrt2 into the base 2.
log sqrt2 (x+1) = 2log2 (x+1)
We'll apply the power rule of logarithms:
2log2 (x+1) = log2 (x+1)^2
We'll re-write the equation:
log2 (x^2+x+1) = log2 (x+1)^2
Since the bases are matching, we'll apply one to one property:
x^2+x+1 = (x+1)^2
We'll expand the square:
x^2 + x + 1 = x^2 + 2x + 1
We'll remove like terms:
x = 2x
x - 2x = 0
-x = 0
x = 0
Since x = 0 > -1, we'll accept the value x = 0.
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