Solve for x log base 2x-5 of 3x^2-7x-15=2

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

Rewrite $\log_{2 x - 5} (3 x^{2} - 7 x - 15) = 2$ 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 $b^{y} = x$ .

${(2 x - 5)}^{2} = 3 x^{2} - 7 x - 15$

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$3 x^{2} - 7 x - 15 = {(2 x - 5)}^{2}$

Simplify ${(2 x - 5)}^{2}$ .

Rewrite.

$3 x^{2} - 7 x - 15 = 0 + 0 + {(2 x - 5)}^{2}$

Rewrite ${(2 x - 5)}^{2}$ as $(2 x - 5) (2 x - 5)$ .

$3 x^{2} - 7 x - 15 = (2 x - 5) (2 x - 5)$

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

Apply the distributive property.

$3 x^{2} - 7 x - 15 = 2 x (2 x - 5) - 5 (2 x - 5)$

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $2$ by $2$ .

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

Multiply $- 5$ by $2$ .

$3 x^{2} - 7 x - 15 = 4 x^{2} - 10 x - 5 (2 x) - 5 \cdot - 5$

Multiply $2$ by $- 5$ .

$3 x^{2} - 7 x - 15 = 4 x^{2} - 10 x - 10 x - 5 \cdot - 5$

Multiply $- 5$ by $- 5$ .

$3 x^{2} - 7 x - 15 = 4 x^{2} - 10 x - 10 x + 25$

Subtract $10 x$ from $- 10 x$ .

$3 x^{2} - 7 x - 15 = 4 x^{2} - 20 x + 25$

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

Subtract $4 x^{2}$ from both sides of the equation.

$3 x^{2} - 7 x - 15 - 4 x^{2} = - 20 x + 25$

Add $20 x$ to both sides of the equation.

$3 x^{2} - 7 x - 15 - 4 x^{2} + 20 x = 25$

Subtract $4 x^{2}$ from $3 x^{2}$ .

$- x^{2} - 7 x - 15 + 20 x = 25$

Add $- 7 x$ and $20 x$ .

$- x^{2} + 13 x - 15 = 25$

Subtract $25$ from both sides of the equation.

$- x^{2} + 13 x - 15 - 25 = 0$

Subtract $25$ from $- 15$ .

$- x^{2} + 13 x - 40 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 13 x - 40$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 13 x - 40 = 0$

Factor $- 1$ out of $13 x$ .

$- (x^{2}) - (- 13 x) - 40 = 0$

Rewrite $- 40$ as $- 1 (40)$ .

$- (x^{2}) - (- 13 x) - 1 \cdot 40 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 13 x)$ .

$- (x^{2} - 13 x) - 1 \cdot 40 = 0$

Factor $- 1$ out of $- (x^{2} - 13 x) - 1 (40)$ .

$- (x^{2} - 13 x + 40) = 0$

Factor.

Factor $x^{2} - 13 x + 40$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $40$ and whose sum is $- 13$ .

$- 8, - 5$

Write the factored form using these integers.

$- ((x - 8) (x - 5)) = 0$

Remove unnecessary parentheses.

$- (x - 8) (x - 5) = 0$

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

$x - 8 = 0$

$x - 5 = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $- (x - 8) (x - 5) = 0$ true.

$x = 8, 5$

Solve for x log base 2x-5 of 3x^2-7x-15=2

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