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

1 Answer
Apr 18, 2018

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.