Solve the equation $\log_2(x-1) = 2 - \log_2(x+2)$ for $x$. We need to find the value(s) of $x$ that satisfy this logarithmic equation. The options provided are: (a) $x = -3$ and $x = 2$, (b) $x = -3$, (c) $x = 2$, (d) $x = 3$ and $x = -2$.

1. Problem Description

Solve the equation $\log_2(x-1) = 2 - \log_2(x+2)$ for $x$. We need to find the value(s) of $x$ that satisfy this logarithmic equation. The options provided are: (a) $x = -3$ and $x = 2$, (b) $x = -3$, (c) $x = 2$, (d) $x = 3$ and $x = -2$.

2. Solution Steps

First, rewrite the equation to isolate the logarithmic terms on one side:

$\log_2(x-1) + \log_2(x+2) = 2$

Using the logarithm product rule, which states that $\log_a(b) + \log_a(c) = \log_a(bc)$, we can combine the two logarithms:

$\log_2((x-1)(x+2)) = 2$

Now, convert the logarithmic equation to an exponential equation using the definition of logarithm: $\log_a(b) = c$ is equivalent to $a^c = b$. In this case, $a = 2$, $b = (x-1)(x+2)$, and $c = 2$. Thus, we have:

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

Simplify the equation:

$(x-1)(x+2) = 4$

$x^2 + 2x - x - 2 = 4$

$x^2 + x - 2 = 4$

$x^2 + x - 6 = 0$

Now, we solve the quadratic equation $x^2 + x - 6 = 0$ by factoring:

$(x+3)(x-2) = 0$

This gives us two possible solutions for $x$:

$x+3 = 0 \implies x = -3$

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

Now we need to check these solutions in the original equation. Remember that the argument of a logarithm must be positive.

If $x = -3$, then we have $\log_2(-3-1) = \log_2(-4)$ and $\log_2(-3+2) = \log_2(-1)$. Since we cannot take the logarithm of a negative number, $x = -3$ is not a valid solution.

If $x = 2$, then we have $\log_2(2-1) = \log_2(1) = 0$ and $\log_2(2+2) = \log_2(4) = 2$. Substituting into the original equation:

$\log_2(2-1) = 2 - \log_2(2+2)$

$\log_2(1) = 2 - \log_2(4)$

$0 = 2 - 2$

$0 = 0$

Since the equation holds true and the arguments of the logarithms are positive, $x = 2$ is a valid solution.

3. Final Answer

x = 2