Solve for x log base 1/2 of x^2+x- log base 1/2 of x^2-x=-1

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

Step 1

Simplify the left side.

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

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

Factor $x$ out of $x^{2} + x$ .

Factor $x$ out of $x^{2}$ .

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

Raise $x$ to the power of $1$ .

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

Factor $x$ out of $x^{1}$ .

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

Factor $x$ out of $x \cdot x + x \cdot 1$ .

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

Factor $x$ out of $x^{2} - x$ .

Factor $x$ out of $x^{2}$ .

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

Factor $x$ out of $- x$ .

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

Factor $x$ out of $x \cdot x + x \cdot - 1$ .

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

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 2

Rewrite $\log_{\frac{1}{2}} (\frac{x + 1}{x - 1}) = - 1$ 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 $b^{y} = x$ .

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

Step 3

Cross multiply to remove the fraction.

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

Step 4

Simplify ${(\frac{1}{2})}^{- 1} (x - 1)$ .

Change the sign of the exponent by rewriting the base as its reciprocal.

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

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

$x + 1 = 2 x - 2$

Step 5

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

Subtract $2 x$ from both sides of the equation.

$x + 1 - 2 x = - 2$

Subtract $2 x$ from $x$ .

$- x + 1 = - 2$

Step 6

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

Subtract $1$ from both sides of the equation.

$- x = - 2 - 1$

Subtract $1$ from $- 2$ .

$- x = - 3$

Step 7

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

Solve for x log base 1/2 of x^2+x- log base 1/2 of x^2-x=-1

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