Solve for x log of 3x+ log of x-2 = log of x-4

$\log (3 x) + \log (x - 2) = \log (x - 4)$

Simplify the left side.

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log (3 x (x - 2)) = \log (x - 4)$

Apply the distributive property.

$\log (3 x \cdot x + 3 x \cdot - 2) = \log (x - 4)$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\log (3 (x \cdot x) + 3 x \cdot - 2) = \log (x - 4)$

Multiply $x$ by $x$ .

$\log (3 x^{2} + 3 x \cdot - 2) = \log (x - 4)$

Multiply $- 2$ by $3$ .

$\log (3 x^{2} - 6 x) = \log (x - 4)$

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$3 x^{2} - 6 x = x - 4$

Solve for $x$ .

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

Subtract $x$ from both sides of the equation.

$3 x^{2} - 6 x - x = - 4$

Subtract $x$ from $- 6 x$ .

$3 x^{2} - 7 x = - 4$

Add $4$ to both sides of the equation.

$3 x^{2} - 7 x + 4 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 4 = 12$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$3 x^{2} - 7 x + 4 = 0$

Rewrite $- 7$ as $- 3$ plus $- 4$

$3 x^{2} + (- 3 - 4) x + 4 = 0$

Apply the distributive property.

$3 x^{2} - 3 x - 4 x + 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 3 x) - 4 x + 4 = 0$

Factor out the greatest common factor (GCF) from each group.

$3 x (x - 1) - 4 (x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$(x - 1) (3 x - 4) = 0$

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

$x - 1 = 0$

$3 x - 4 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $3 x - 4$ equal to $0$ and solve for $x$ .

Set $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Solve $3 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$3 x = 4$

Divide each term in $3 x = 4$ by $3$ and simplify.

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Divide $x$ by $1$ .

$x = \frac{4}{3}$

The final solution is all the values that make $(x - 1) (3 x - 4) = 0$ true.

$x = 1, \frac{4}{3}$

Exclude the solutions that do not make $\log (3 x) + \log (x - 2) = \log (x - 4)$ true.

$No solution$

Solve for x log of 3x+ log of x-2 = log of x-4

Related Questions

Support LCMGCF.com by disabling your adblocker.