Question

Log2(x+1)=log4(x^2-x+4) What is the value of x?

Answer 1

`x=1`

Explanation:

`log_2(x+1)=log_4(x^2-x+4)`

Lets start by converting the `log_4` term to a `log_2` term. We can do this by using the change-of-base formula.

`log_ax=(log_bx)/(log_ba)`

So we can rewrite our expression as

`log_2(x+1)=(log_2(x^2-x+4))/(log_2(4))`

But `log_2(4)=2` so,

`log_2(x+1)=(log_2(x^2-x+4))/2`

Multiply both sides of this equation by 2.

`2log_2(x+1)=log_2(x^2-x+4)`

Now we can use the property of logarithms that says `alogx=logx^a`. This lets us rewrite our equation as

`log_2(x+1)^2=log_2(x^2-x+4)`.

If the logs are equal, then their arguments must be equal.

`(x+1)^2=x^2-x+4`

Expand the squared binomial on the left-hand side of this equation.

`x^2+2x+1=x^2-x+4`

Combine like terms.

`3x=3`

Divide both sides by 3.

`x=1`

Check to make sure this makes sense by plugging this value for `x` into the original equation.

`log_2(1+1)` =?= `log_4(1^2-1+4)`

`log_2(2)` =?= `log_4(4)`

`1=1`

Yup.