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

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### 3 Answers

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.**

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.**