Solve for x log base 4 of 3x^2-4- log base 4 of 4x+2=1

$\log_{4} (3 x^{2} - 4) - \log_{4} (4 x + 2) = 1$

Simplify the left side.

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

$\log_{4} (\frac{3 x^{2} - 4}{4 x + 2}) = 1$

Factor $2$ out of $4 x + 2$ .

Factor $2$ out of $4 x$ .

$\log_{4} (\frac{3 x^{2} - 4}{2 (2 x) + 2}) = 1$

Factor $2$ out of $2$ .

$\log_{4} (\frac{3 x^{2} - 4}{2 (2 x) + 2 (1)}) = 1$

Factor $2$ out of $2 (2 x) + 2 (1)$ .

$\log_{4} (\frac{3 x^{2} - 4}{2 (2 x + 1)}) = 1$

Rewrite $\log_{4} (\frac{3 x^{2} - 4}{2 (2 x + 1)}) = 1$ 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 $b^{y} = x$ .

$4^{1} = \frac{3 x^{2} - 4}{2 (2 x + 1)}$

Cross multiply to remove the fraction.

$3 x^{2} - 4 = 4 (2 (2 x + 1))$

Simplify $4 (2 (2 x + 1))$ .

Apply the distributive property.

$3 x^{2} - 4 = 4 (2 (2 x) + 2 \cdot 1)$

Multiply.

Multiply $2$ by $2$ .

$3 x^{2} - 4 = 4 (4 x + 2 \cdot 1)$

Multiply $2$ by $1$ .

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

Apply the distributive property.

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

Multiply.

Multiply $4$ by $4$ .

$3 x^{2} - 4 = 16 x + 4 \cdot 2$

Multiply $4$ by $2$ .

$3 x^{2} - 4 = 16 x + 8$

Subtract $16 x$ from both sides of the equation.

$3 x^{2} - 4 - 16 x = 8$

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

Add $4$ to both sides of the equation.

$3 x^{2} - 16 x = 8 + 4$

Add $8$ and $4$ .

$3 x^{2} - 16 x = 12$

Subtract $12$ from both sides of the equation.

$3 x^{2} - 16 x - 12 = 0$

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 $a \cdot c = 3 \cdot - 12 = - 36$ and whose sum is $b = - 16$ .

Factor $- 16$ out of $- 16 x$ .

$3 x^{2} - 16 x - 12 = 0$

Rewrite $- 16$ as $2$ plus $- 18$

$3 x^{2} + (2 - 18) x - 12 = 0$

Apply the distributive property.

$3 x^{2} + 2 x - 18 x - 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} + 2 x) - 18 x - 12 = 0$

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

$x (3 x + 2) - 6 (3 x + 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x + 2$ .

$(3 x + 2) (x - 6) = 0$

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

$3 x + 2 = 0$

$x - 6 = 0$

Set $3 x + 2$ equal to $0$ and solve for $x$ .

Set $3 x + 2$ equal to $0$ .

$3 x + 2 = 0$

Solve $3 x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Divide each term in $3 x = - 2$ by $3$ and simplify.

Divide each term in $3 x = - 2$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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.

$x = 6$

The final solution is all the values that make $(3 x + 2) (x - 6) = 0$ true.

$x = - \frac{2}{3}, 6$

Exclude the solutions that do not make $\log_{4} (3 x^{2} - 4) - \log_{4} (4 x + 2) = 1$ true.

$x = 6$

Solve for x log base 4 of 3x^2-4- log base 4 of 4x+2=1

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