Solve for x log base 4 of x+5+ log base 4 of 3x-7=2

$\log_{4} (x + 5) + \log_{4} (3 x - 7) = 2$

Step 1

Simplify the left side.

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

$\log_{4} ((x + 5) (3 x - 7)) = 2$

Expand $(x + 5) (3 x - 7)$ using the FOIL Method.

Apply the distributive property.

$\log_{4} (x (3 x - 7) + 5 (3 x - 7)) = 2$

Apply the distributive property.

$\log_{4} (x (3 x) + x \cdot - 7 + 5 (3 x - 7)) = 2$

Apply the distributive property.

$\log_{4} (x (3 x) + x \cdot - 7 + 5 (3 x) + 5 \cdot - 7) = 2$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$\log_{4} (3 x \cdot x + x \cdot - 7 + 5 (3 x) + 5 \cdot - 7) = 2$

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

Move $x$ .

$\log_{4} (3 (x \cdot x) + x \cdot - 7 + 5 (3 x) + 5 \cdot - 7) = 2$

Multiply $x$ by $x$ .

$\log_{4} (3 x^{2} + x \cdot - 7 + 5 (3 x) + 5 \cdot - 7) = 2$

Move $- 7$ to the left of $x$ .

$\log_{4} (3 x^{2} - 7 \cdot x + 5 (3 x) + 5 \cdot - 7) = 2$

Multiply $3$ by $5$ .

$\log_{4} (3 x^{2} - 7 x + 15 x + 5 \cdot - 7) = 2$

Multiply $5$ by $- 7$ .

$\log_{4} (3 x^{2} - 7 x + 15 x - 35) = 2$

Add $- 7 x$ and $15 x$ .

$\log_{4} (3 x^{2} + 8 x - 35) = 2$

Step 2

Rewrite $\log_{4} (3 x^{2} + 8 x - 35) = 2$ 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 $b^{y} = x$ .

$4^{2} = 3 x^{2} + 8 x - 35$

Step 3

Solve for $x$ .

Rewrite the equation as $3 x^{2} + 8 x - 35 = 4^{2}$ .

$3 x^{2} + 8 x - 35 = 4^{2}$

Raise $4$ to the power of $2$ .

$3 x^{2} + 8 x - 35 = 16$

Subtract $16$ from both sides of the equation.

$3 x^{2} + 8 x - 35 - 16 = 0$

Subtract $16$ from $- 35$ .

$3 x^{2} + 8 x - 51 = 0$

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 $a \cdot c = 3 \cdot - 51 = - 153$ and whose sum is $b = 8$ .

Factor $8$ out of $8 x$ .

$3 x^{2} + 8 (x) - 51 = 0$

Rewrite $8$ as $- 9$ plus $17$

$3 x^{2} + (- 9 + 17) x - 51 = 0$

Apply the distributive property.

$3 x^{2} - 9 x + 17 x - 51 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 9 x) + 17 x - 51 = 0$

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

$3 x (x - 3) + 17 (x - 3) = 0$

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

$(x - 3) (3 x + 17) = 0$

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

$x - 3 = 0$

$3 x + 17 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $3 x + 17$ equal to $0$ and solve for $x$ .

Set $3 x + 17$ equal to $0$ .

$3 x + 17 = 0$

Solve $3 x + 17 = 0$ for $x$ .

Subtract $17$ from both sides of the equation.

$3 x = - 17$

Divide each term in $3 x = - 17$ by $3$ and simplify.

Divide each term in $3 x = - 17$ by $3$ .

$\frac{3 x}{3} = \frac{- 17}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 17}{3}$

Divide $x$ by $1$ .

$x = \frac{- 17}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{17}{3}$

The final solution is all the values that make $(x - 3) (3 x + 17) = 0$ true.

$x = 3, - \frac{17}{3}$

Step 4

Exclude the solutions that do not make $\log_{4} (x + 5) + \log_{4} (3 x - 7) = 2$ true.

$x = 3$

Solve for x log base 4 of x+5+ log base 4 of 3x-7=2

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