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

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

Move all the terms containing a logarithm to the left side of the equation.

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

Simplify the left side.

Simplify $2 \log_{3} (x) - \log_{3} (x - 2)$ .

Simplify $2 \log_{3} (x)$ by moving $2$ inside the logarithm.

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

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

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

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

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

Cross multiply to remove the fraction.

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

Simplify $3^{2} (x - 2)$ .

Raise $3$ to the power of $2$ .

$x^{2} = 9 (x - 2)$

Apply the distributive property.

$x^{2} = 9 x + 9 \cdot - 2$

Multiply $9$ by $- 2$ .

$x^{2} = 9 x - 18$

Subtract $9 x$ from both sides of the equation.

$x^{2} - 9 x = - 18$

Factor $x$ out of $x^{2} - 9 x$ .

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

$x \cdot x - 9 x = - 18$

Factor $x$ out of $- 9 x$ .

$x \cdot x + x \cdot - 9 = - 18$

Factor $x$ out of $x \cdot x + x \cdot - 9$ .

$x (x - 9) = - 18$

Simplify $x (x - 9)$ .

Apply the distributive property.

$x \cdot x + x \cdot - 9 = - 18$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 9 = - 18$

Move $- 9$ to the left of $x$ .

$x^{2} - 9 x = - 18$

Add $18$ to both sides of the equation.

$x^{2} - 9 x + 18 = 0$

Factor $x^{2} - 9 x + 18$ 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 $18$ and whose sum is $- 9$ .

$- 6, - 3$

Write the factored form using these integers.

$(x - 6) (x - 3) = 0$

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

$x - 6 = 0$

$x - 3 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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.

$x = 3$

The final solution is all the values that make $(x - 6) (x - 3) = 0$ true.

$x = 6, 3$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.