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

Apr 18, 2018

$x = 1$

#### Explanation:

${\log}_{2} \left(x + 1\right) = {\log}_{4} \left({x}^{2} - x + 4\right)$

Lets start by converting the ${\log}_{4}$ term to a ${\log}_{2}$ term. We can do this by using the change-of-base formula.

${\log}_{a} x = \frac{{\log}_{b} x}{{\log}_{b} a}$

So we can rewrite our expression as

${\log}_{2} \left(x + 1\right) = \frac{{\log}_{2} \left({x}^{2} - x + 4\right)}{{\log}_{2} \left(4\right)}$

But ${\log}_{2} \left(4\right) = 2$ so,

${\log}_{2} \left(x + 1\right) = \frac{{\log}_{2} \left({x}^{2} - x + 4\right)}{2}$

Multiply both sides of this equation by 2.

$2 {\log}_{2} \left(x + 1\right) = {\log}_{2} \left({x}^{2} - x + 4\right)$

Now we can use the property of logarithms that says $a \log x = \log {x}^{a}$. This lets us rewrite our equation as

${\log}_{2} {\left(x + 1\right)}^{2} = {\log}_{2} \left({x}^{2} - x + 4\right)$.

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

${\left(x + 1\right)}^{2} = {x}^{2} - x + 4$

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

${x}^{2} + 2 x + 1 = {x}^{2} - x + 4$

Combine like terms.

$3 x = 3$

Divide both sides by 3.

$x = 1$

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

${\log}_{2} \left(1 + 1\right)$ =?= ${\log}_{4} \left({1}^{2} - 1 + 4\right)$

${\log}_{2} \left(2\right)$ =?= ${\log}_{4} \left(4\right)$

$1 = 1$

Yup.