Solve for x log base 3 of x^2-2x=1

$\log_{3} (x^{2} - 2 x) = 1$

Rewrite $\log_{3} (x^{2} - 2 x) = 1$ 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 $b^{y} = x$ .

$3^{1} = x^{2} - 2 x$

Solve for $x$ .

Rewrite the equation as $x^{2} - 2 x = 3^{1}$ .

$x^{2} - 2 x = 3$

Subtract $3$ from both sides of the equation.

$x^{2} - 2 x - 3 = 0$

Factor $x^{2} - 2 x - 3$ 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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$(x - 3) (x + 1) = 0$

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

$x - 3 = 0$

$x + 1 = 0$

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 $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Solve for x log base 3 of x^2-2x=1

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