Solve for x log base 4 of 4x^2+7- log base 4 of 2x+1=1

$\log_{4} (4 x^{2} + 7) - \log_{4} (2 x + 1) = 1$

Step 1

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

$\log_{4} (\frac{4 x^{2} + 7}{2 x + 1}) = 1$

Step 2

Rewrite $\log_{4} (\frac{4 x^{2} + 7}{2 x + 1}) = 1$ 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 $b^{y} = x$ .

$4^{1} = \frac{4 x^{2} + 7}{2 x + 1}$

Step 3

Cross multiply to remove the fraction.

$4 x^{2} + 7 = 4 (2 x + 1)$

Step 4

Simplify $4 (2 x + 1)$ .

Apply the distributive property.

$4 x^{2} + 7 = 4 (2 x) + 4 \cdot 1$

Multiply.

Multiply $2$ by $4$ .

$4 x^{2} + 7 = 8 x + 4 \cdot 1$

Multiply $4$ by $1$ .

$4 x^{2} + 7 = 8 x + 4$

Step 5

Subtract $8 x$ from both sides of the equation.

$4 x^{2} + 7 - 8 x = 4$

Step 6

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

Subtract $7$ from both sides of the equation.

$4 x^{2} - 8 x = 4 - 7$

Subtract $7$ from $4$ .

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

Step 7

Add $3$ to both sides of the equation.

$4 x^{2} - 8 x + 3 = 0$

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 $a \cdot c = 4 \cdot 3 = 12$ and whose sum is $b = - 8$ .

Factor $- 8$ out of $- 8 x$ .

$4 x^{2} - 8 x + 3 = 0$

Rewrite $- 8$ as $- 2$ plus $- 6$

$4 x^{2} + (- 2 - 6) x + 3 = 0$

Apply the distributive property.

$4 x^{2} - 2 x - 6 x + 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 x^{2} - 2 x) - 6 x + 3 = 0$

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

$2 x (2 x - 1) - 3 (2 x - 1) = 0$

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

$(2 x - 1) (2 x - 3) = 0$

Step 9

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

$2 x - 1 = 0$

$2 x - 3 = 0$

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 $2 x - 3$ equal to $0$ and solve for $x$ .

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

$2 x - 3 = 0$

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

Add $3$ to both sides of the equation.

$2 x = 3$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 12

The final solution is all the values that make $(2 x - 1) (2 x - 3) = 0$ true.

$x = \frac{1}{2}, \frac{3}{2}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{2}, \frac{3}{2}$

Decimal Form:

$x = 0.5, 1.5$

Mixed Number Form:

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

Solve for x log base 4 of 4x^2+7- log base 4 of 2x+1=1

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