Solve for x log base x of 3x^2-10x+12=2

$\log_{x} (3 x^{2} - 10 x + 12) = 2$

Rewrite $\log_{x} (3 x^{2} - 10 x + 12) = 2$ 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 $b^{y} = x$ .

$x^{2} = 3 x^{2} - 10 x + 12$

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$3 x^{2} - 10 x + 12 = x^{2}$

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

Subtract $x^{2}$ from both sides of the equation.

$3 x^{2} - 10 x + 12 - x^{2} = 0$

Subtract $x^{2}$ from $3 x^{2}$ .

$2 x^{2} - 10 x + 12 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 10 x + 12$ .

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

$2 (x^{2}) - 10 x + 12 = 0$

Factor $2$ out of $- 10 x$ .

$2 (x^{2}) + 2 (- 5 x) + 12 = 0$

Factor $2$ out of $12$ .

$2 x^{2} + 2 (- 5 x) + 2 \cdot 6 = 0$

Factor $2$ out of $2 x^{2} + 2 (- 5 x)$ .

$2 (x^{2} - 5 x) + 2 \cdot 6 = 0$

Factor $2$ out of $2 (x^{2} - 5 x) + 2 \cdot 6$ .

$2 (x^{2} - 5 x + 6) = 0$

Factor.

Factor $x^{2} - 5 x + 6$ 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 $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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 $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 3, 2$

Solve for x log base x of 3x^2-10x+12=2

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