Solve the next three questions

Since $8 = 2^3$, we can rewrite the equation:

$$2^{x+1} = (2^3)^{x-1}$$

$$2^{x+1} = 2^{3(x-1)}$$ $$2^{x+1} = 2^{3x-3}$$

$$1+3 = 3x-x$$ $$4 = 2x$$ $$x = \frac{4}{2}$$ $$x = 2$$

The solution is:

$$\boxed{x = 2}$$

Recall that $\log_b A = C$ is equivalent to $b^C = A$.

$$x^2 - 1 = 2^3$$ $$x^2 - 1 = 8$$

$$x^2 = 8 + 1$$ $$x^2 = 9$$ $$x = \pm\sqrt{9}$$ $$x = \pm 3$$

The arguments of a logarithm must be positive.
For $\log_2(x-1)$: $x-1 > 0 \implies x > 1$.
For $\log_2(x+1)$: $x+1 > 0 \implies x > -1$.
Both conditions require $x > 1$.
If $x = 3$: $3 > 1$, so $x=3$ is a valid solution.
If $x = -3$: $-3 \ngtr 1$, so $x=-3$ is an extraneous solution.

The solution is:

$$\boxed{x = 3}$$

$$(\sqrt{x+2})^2 = x^2$$ $$x+2 = x^2$$

We need two numbers that multiply to -2 and add to -1. These numbers are -2 and 1.

$$(x-2)(x+1) = 0$$

$$x-2 = 0 \implies x = 2$$ $$x+1 = 0 \implies x = -1$$

When squaring both sides of an equation, extraneous solutions can be introduced. We must check both potential solutions in the original equation $\sqrt{x+2} = x$.

Check $x=2$:

$$\sqrt{2+2} = 2$$ $$\sqrt{4} = 2$$ $$2 = 2$$

This is true, so $x=2$ is a valid solution.

Check $x=-1$:

$$\sqrt{-1+2} = -1$$ $$\sqrt{1} = -1$$ $$1 = -1$$

This is false, so $x=-1$ is an extraneous solution.

The solution is:

$$\boxed{x = 2}$$