Solve the logarithmic equation: $log_2 (6 - x) = 3 - log_2 x$

Analysis

To solve the equation $\log_2 (6 - x) = 3 - \log_2 x$, we need to use the properties of logarithms to combine the terms and convert the logarithmic equation into an algebraic one.

Concept

1. Logarithm Addition Rule: $\log_b(A) + \log_b(B) = \log_b(A \times B)$.
2. Definition of Logarithm: If $\log_b(x) = y$, then $x = b^y$.
3. Domain Constraint: The argument of a logarithm must always be positive (i.e., $6 - x > 0$ and $x > 0$).

Step-by-Step Solution

Step 1: Rearrange the equation Move the $\log_2 x$ term to the left side:

Step 2: Apply the addition rule Combine the logarithms on the left:

Step 3: Convert to exponential form Using the definition $\log_b(A) = y \implies A = b^y$:

Step 4: Solve the quadratic equation Rearrange all terms to one side to set the equation to zero:

Factor the quadratic equation: Find two numbers that multiply to $+8$ and add to $-6$. These numbers are $-4$ and $-2$.

Setting each factor to zero: $x - 4 = 0 \implies x = 4$ $x - 2 = 0 \implies x = 2$

Step 5: Verify the solutions Check if these values satisfy the domain constraints ($x > 0$ and $6-x > 0$):

• If $x = 4$: $6 - 4 = 2$ (positive), $x = 4$ (positive). Valid.
• If $x = 2$: $6 - 2 = 4$ (positive), $x = 2$ (positive). Valid.

Conclusion

The solutions are $x = 4$ or $x = 2$. This matches option A.

Exam Tip

Always check your answers in the original logarithmic equation. If you had obtained a negative value for $x$, you would have to reject it, because you cannot take the logarithm of a negative number or zero. In JAMB, they often include "distractor" answers that are mathematically correct for the quadratic but invalid for the logarithm—always double-check your domain!