Solve for x log base 7 of x^2-5- log base 7 of x+1=0

$\log_{7} (x^{2} - 5) - \log_{7} (x + 1) = 0$

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

$\log_{7} (\frac{x^{2} - 5}{x + 1}) = 0$

Rewrite $\log_{7} (\frac{x^{2} - 5}{x + 1}) = 0$ 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 $b^{y} = x$ .

$7^{0} = \frac{x^{2} - 5}{x + 1}$

Cross multiply to remove the fraction.

$x^{2} - 5 = 7^{0} (x + 1)$

Simplify $7^{0} (x + 1)$ .

Anything raised to $0$ is $1$ .

$x^{2} - 5 = 1 (x + 1)$

Multiply $x + 1$ by $1$ .

$x^{2} - 5 = x + 1$

Subtract $x$ from both sides of the equation.

$x^{2} - 5 - x = 1$

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

Add $5$ to both sides of the equation.

$x^{2} - x = 1 + 5$

Add $1$ and $5$ .

$x^{2} - x = 6$

Subtract $6$ from both sides of the equation.

$x^{2} - x - 6 = 0$

Factor $x^{2} - x - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$(x - 3) (x + 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x + 2 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 3) (x + 2) = 0$ true.

$x = 3, - 2$

Exclude the solutions that do not make $\log_{7} (x^{2} - 5) - \log_{7} (x + 1) = 0$ true.

$x = 3$

Solve for x log base 7 of x^2-5- log base 7 of x+1=0

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