Solving the Logarithmic Equation for $x$

This solution explains how to find the value of $x$ in the given logarithmic equation: $log_3(x-2) + log_3(x+4) = 3$. We will use properties of logarithms and algebraic methods to solve for $x$.

Understanding Logarithm Properties

To solve this equation, we need to use a key property of logarithms, which states that the sum of two logarithms with the same base can be combined into a single logarithm by multiplying their arguments:

$$log_b(M) + log_b(N) = log_b(M \times N)$$

We also use the definition of a logarithm: if $log_b(A) = C$, then $b^C = A$.

Additionally, remember that the argument of a logarithm must always be positive. This means for $log_b(y)$, we must have $y > 0$.

Step-by-Step Solution

Step 1: Combine the Logarithms

Apply the logarithm property $log_b(M) + log_b(N) = log_b(M \times N)$ to the left side of the equation:

$$log_3(x-2) + log_3(x+4) = 3$$

$$log_3((x-2)(x+4)) = 3$$

Step 2: Convert to Exponential Form

Now, convert the logarithmic equation to its equivalent exponential form using the definition $log_b(A) = C \implies b^C = A$. Here, the base $b=3$, the argument $A=(x-2)(x+4)$, and the result $C=3$.

$$ (x-2)(x+4) = 3^3 $$

Step 3: Simplify and Solve the Quadratic Equation

Expand the left side and calculate the right side:

$$ x^2 + 4x - 2x - 8 = 27 $$

Combine like terms:

$$ x^2 + 2x - 8 = 27 $$

Move all terms to one side to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$$ x^2 + 2x - 8 - 27 = 0 $$

$$ x^2 + 2x - 35 = 0 $$

Factor the quadratic equation. We need two numbers that multiply to -35 and add to 2. These numbers are 7 and -5.

$$ (x+7)(x-5) = 0 $$

Set each factor equal to zero to find the potential solutions:

• $x + 7 = 0 \implies x = -7$
• $x - 5 = 0 \implies x = 5$

So, the potential solutions are $x = -7$ and $x = 5$.

Step 4: Check for Domain Validity

It's crucial to check if these potential solutions satisfy the domain requirements of the original logarithmic equation. The arguments of the logarithms must be greater than zero:

• For $log_3(x-2)$, we require $x-2 > 0$, which means $x > 2$.
• For $log_3(x+4)$, we require $x+4 > 0$, which means $x > -4$.

For the original equation to be defined, both conditions must hold true. Therefore, we must have $x > 2$.

Step 5: Validate the Potential Solutions

Now, we check our potential solutions against the condition $x > 2$:

• Check $x = -7$: Is $-7 > 2$? No. This solution is not valid because it violates the domain requirement ($x-2$ would be negative).
• Check $x = 5$: Is $5 > 2$? Yes. This solution is valid because it satisfies the domain requirement ($x-2 = 3 > 0$ and $x+4 = 9 > 0$).

Conclusion

After checking the potential solutions against the domain restrictions of the original logarithmic equation, we find that only $x = 5$ is a valid solution.