Solve for x log base 4 of x+2- log base 4 of x-1=1

$\log_{4} (x + 2) - \log_{4} (x - 1) = 1$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{4} (\frac{x + 2}{x - 1}) = 1$

Rewrite $\log_{4} (\frac{x + 2}{x - 1}) = 1$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ $\neq$ $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

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

Cross multiply to remove the fraction.

$x + 2 = 4 (x - 1)$

Simplify $4 (x - 1)$ .

Apply the distributive property.

$x + 2 = 4 x + 4 \cdot - 1$

Multiply $4$ by $- 1$ .

$x + 2 = 4 x - 4$

Move all terms containing $x$ to the left side of the equation.

Subtract $4 x$ from both sides of the equation.

$x + 2 - 4 x = - 4$

Subtract $4 x$ from $x$ .

$- 3 x + 2 = - 4$

Move all terms not containing $x$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$- 3 x = - 4 - 2$

Subtract $2$ from $- 4$ .

$- 3 x = - 6$

Divide each term in $- 3 x = - 6$ by $- 3$ and simplify.

Divide each term in $- 3 x = - 6$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{- 6}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{- 6}{- 3}$

Divide $x$ by $1$ .

$x = \frac{- 6}{- 3}$

Simplify the right side.

Divide $- 6$ by $- 3$ .

$x = 2$

Solve for x log base 4 of x+2- log base 4 of x-1=1

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