Solve for x log base 7 of 2x^2+4- log base 7 of 3x+3=0

$\log_{7} (2 x^{2} + 4) - \log_{7} (3 x + 3) = 0$

Step 1

Simplify the left side.

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

$\log_{7} (\frac{2 x^{2} + 4}{3 x + 3}) = 0$

Factor $2$ out of $2 x^{2} + 4$ .

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

$\log_{7} (\frac{2 (x^{2}) + 4}{3 x + 3}) = 0$

Factor $2$ out of $4$ .

$\log_{7} (\frac{2 x^{2} + 2 \cdot 2}{3 x + 3}) = 0$

Factor $2$ out of $2 x^{2} + 2 \cdot 2$ .

$\log_{7} (\frac{2 (x^{2} + 2)}{3 x + 3}) = 0$

Factor $3$ out of $3 x + 3$ .

Factor $3$ out of $3 x$ .

$\log_{7} (\frac{2 (x^{2} + 2)}{3 (x) + 3}) = 0$

Factor $3$ out of $3$ .

$\log_{7} (\frac{2 (x^{2} + 2)}{3 (x) + 3 (1)}) = 0$

Factor $3$ out of $3 (x) + 3 (1)$ .

$\log_{7} (\frac{2 (x^{2} + 2)}{3 (x + 1)}) = 0$

Step 2

Rewrite $\log_{7} (\frac{2 (x^{2} + 2)}{3 (x + 1)}) = 0$ 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 $b^{y} = x$ .

$7^{0} = \frac{2 (x^{2} + 2)}{3 (x + 1)}$

Step 3

Cross multiply to remove the fraction.

$2 (x^{2} + 2) = 7^{0} (3 (x + 1))$

Step 4

Simplify $7^{0} (3 (x + 1))$ .

Simplify the expression.

Anything raised to $0$ is $1$ .

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

Multiply $3 (x + 1)$ by $1$ .

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

Apply the distributive property.

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

Multiply $3$ by $1$ .

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

Step 5

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

Subtract $3 x$ from both sides of the equation.

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

Simplify each term.

Apply the distributive property.

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

Multiply $2$ by $2$ .

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

Step 6

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

Subtract $4$ from both sides of the equation.

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

Subtract $4$ from $3$ .

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

Step 7

Add $1$ to both sides of the equation.

$2 x^{2} - 3 x + 1 = 0$

Step 8

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 $a \cdot c = 2 \cdot 1 = 2$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 x$ .

$2 x^{2} - 3 x + 1 = 0$

Rewrite $- 3$ as $- 1$ plus $- 2$

$2 x^{2} + (- 1 - 2) x + 1 = 0$

Apply the distributive property.

$2 x^{2} - 1 x - 2 x + 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) - 2 x + 1 = 0$

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

$x (2 x - 1) - (2 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x - 1) = 0$

Step 9

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

$2 x - 1 = 0$

$x - 1 = 0$

Step 10

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 11

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.

$x = 1$

Step 12

The final solution is all the values that make $(2 x - 1) (x - 1) = 0$ true.

$x = \frac{1}{2}, 1$

Step 13

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{2}, 1$

Decimal Form:

$x = 0.5, 1$

Solve for x log base 7 of 2x^2+4- log base 7 of 3x+3=0

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