Find the value(s) of x in the following: a) 3^x imes 9^-x = (27)^x+2 b) 7(6x2 − x − 1) = 0

$$3^x \times (3^2)^{-x} = (3^3)^{x+2}$$ $$3^x \times 3^{-2x} = 3^{3(x+2)}$$

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

$$-x = 3x + 6$$ $$-4x = 6$$ $$x = -\frac{6}{4}$$ $$x = -\frac{3}{2}$$

The value of $x$ is $\boxed{-\frac{3}{2}}$.

if $\log_b A = C$, then $b^C = A$.

$$6x^2 - x - 1 = 7^0$$ $$6x^2 - x - 1 = 1$$

$$6x^2 - x - 2 = 0$$

We can factor this quadratic equation. We need two numbers that multiply to $6 \times (-2) = -12$ and add to $-1$. These numbers are $-4$ and 3.

$$6x^2 - 4x + 3x - 2 = 0$$ $$2x(3x - 2) + 1(3x - 2) = 0$$ $$(2x + 1)(3x - 2) = 0$$

This gives two possible solutions for $x$:

$$2x + 1 = 0 \implies x = -\frac{1}{2}$$ $$3x - 2 = 0 \implies x = \frac{2}{3}$$

The argument of the logarithm must be positive.
For $x = -\frac{1}{2}$:
$6\left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right) - 1 = 6\left(\frac{1}{4}\right) + \frac{1}{2} - 1 = \frac{3}{2} + \frac{1}{2} - 1 = \frac{4}{2} - 1 = 2 - 1 = 1$. Since $1 > 0$, $x = -\frac{1}{2}$ is a valid solution.

For $x = \frac{2}{3}$:
$6\left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right) - 1 = 6\left(\frac{4}{9}\right) - \frac{2}{3} - 1 = \frac{24}{9} - \frac{2}{3} - 1 = \frac{8}{3} - \frac{2}{3} - 1 = \frac{6}{3} - 1 = 2 - 1 = 1$. Since $1 > 0$, $x = \frac{2}{3}$ is a valid solution.

The values of $x