Solve for x log base 4 of 3x+5+ log base 4 of 2x+3=0

$\log_{4} (3 x + 5) + \log_{4} (2 x + 3) = 0$

Step 1

Simplify the left side.

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

$\log_{4} ((3 x + 5) (2 x + 3)) = 0$

Expand $(3 x + 5) (2 x + 3)$ using the FOIL Method.

Apply the distributive property.

$\log_{4} (3 x (2 x + 3) + 5 (2 x + 3)) = 0$

Apply the distributive property.

$\log_{4} (3 x (2 x) + 3 x \cdot 3 + 5 (2 x + 3)) = 0$

Apply the distributive property.

$\log_{4} (3 x (2 x) + 3 x \cdot 3 + 5 (2 x) + 5 \cdot 3) = 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$\log_{4} (3 \cdot (2 x \cdot x) + 3 x \cdot 3 + 5 (2 x) + 5 \cdot 3) = 0$

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

Move $x$ .

$\log_{4} (3 \cdot (2 (x \cdot x)) + 3 x \cdot 3 + 5 (2 x) + 5 \cdot 3) = 0$

Multiply $x$ by $x$ .

$\log_{4} (3 \cdot (2 x^{2}) + 3 x \cdot 3 + 5 (2 x) + 5 \cdot 3) = 0$

Multiply $3$ by $2$ .

$\log_{4} (6 x^{2} + 3 x \cdot 3 + 5 (2 x) + 5 \cdot 3) = 0$

Multiply $3$ by $3$ .

$\log_{4} (6 x^{2} + 9 x + 5 (2 x) + 5 \cdot 3) = 0$

Multiply $2$ by $5$ .

$\log_{4} (6 x^{2} + 9 x + 10 x + 5 \cdot 3) = 0$

Multiply $5$ by $3$ .

$\log_{4} (6 x^{2} + 9 x + 10 x + 15) = 0$

Add $9 x$ and $10 x$ .

$\log_{4} (6 x^{2} + 19 x + 15) = 0$

Step 2

Rewrite $\log_{4} (6 x^{2} + 19 x + 15) = 0$ 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 $b^{y} = x$ .

$4^{0} = 6 x^{2} + 19 x + 15$

Step 3

Solve for $x$ .

Rewrite the equation as $6 x^{2} + 19 x + 15 = 4^{0}$ .

$6 x^{2} + 19 x + 15 = 4^{0}$

Anything raised to $0$ is $1$ .

$6 x^{2} + 19 x + 15 = 1$

Subtract $1$ from both sides of the equation.

$6 x^{2} + 19 x + 15 - 1 = 0$

Subtract $1$ from $15$ .

$6 x^{2} + 19 x + 14 = 0$

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 $a \cdot c = 6 \cdot 14 = 84$ and whose sum is $b = 19$ .

Factor $19$ out of $19 x$ .

$6 x^{2} + 19 (x) + 14 = 0$

Rewrite $19$ as $7$ plus $12$

$6 x^{2} + (7 + 12) x + 14 = 0$

Apply the distributive property.

$6 x^{2} + 7 x + 12 x + 14 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(6 x^{2} + 7 x) + 12 x + 14 = 0$

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

$x (6 x + 7) + 2 (6 x + 7) = 0$

Factor the polynomial by factoring out the greatest common factor, $6 x + 7$ .

$(6 x + 7) (x + 2) = 0$

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

$6 x + 7 = 0$

$x + 2 = 0$

Set $6 x + 7$ equal to $0$ and solve for $x$ .

Set $6 x + 7$ equal to $0$ .

$6 x + 7 = 0$

Solve $6 x + 7 = 0$ for $x$ .

Subtract $7$ from both sides of the equation.

$6 x = - 7$

Divide each term in $6 x = - 7$ by $6$ and simplify.

Divide each term in $6 x = - 7$ by $6$ .

$\frac{6 x}{6} = \frac{- 7}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{- 7}{6}$

Divide $x$ by $1$ .

$x = \frac{- 7}{6}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{7}{6}$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(6 x + 7) (x + 2) = 0$ true.

$x = - \frac{7}{6}, - 2$

Step 4

Exclude the solutions that do not make $\log_{4} (3 x + 5) + \log_{4} (2 x + 3) = 0$ true.

$x = - \frac{7}{6}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{7}{6}$

Decimal Form:

$x = - 1.1\overline{6}$

Mixed Number Form:

$x = - 1 \frac{1}{6}$

Solve for x log base 4 of 3x+5+ log base 4 of 2x+3=0

Related Questions

Support LCMGCF.com by disabling your adblocker.