Solve Log(x²−5x−6)=3

The given equation is $\log_2(x^2 - 5x - 6) = 3$.
Using the definition of logarithm, $\log_b a = c \iff b^c = a$, we can rewrite the equation as:

$$2^3 = x^2 - 5x - 6$$

Calculate the value of $2^3$:

$$8 = x^2 - 5x - 6$$

Subtract 8 from both sides of the equation:

$$x^2 - 5x - 6 - 8 = 0$$ $$x^2 - 5x - 14 = 0$$

We need two numbers that multiply to -14 and add up to -5. These numbers are -7 and 2.

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

Set each factor equal to zero:

$$x - 7 = 0 \implies x = 7$$ $$x + 2 = 0 \implies x = -2$$

The argument of a logarithm must be positive, so $x^2 - 5x - 6 > 0$.

For $x = 7$:

$$(7)^2 - 5(7) - 6 = 49 - 35 - 6 = 14 - 6 = 8$$

Since $8 > 0$, $x=7$ is a valid solution.

For $x = -2$:

$$(-2)^2 - 5(-2) - 6 = 4 + 10 - 6 = 14 - 6 = 8$$

Since $8 > 0$, $x=-2$ is a valid solution.

The solutions are $x=7$ and $x=-2$.

$$x = 7, x = -2$$