Question

If \(\frac{1}{2}\log x + \frac{1}{2}\log y + \log 2 = \log(x+y)\), then:

• A. \(x = -y\)
• B. \(x = y + 1\)
• C. \(x = y\)
• D. \(y = x + 1\)

Hint:

Combine the logs into one, drop the log from both sides, then complete the square.

Answer 1

The Correct Option is C

Step 1: Rewrite the equation without expanding logs into single fractions first.
$\frac{1}{2}\log x + \frac{1}{2}\log y = \frac{1}{2}\log(xy)$ by the power and product rules. So the full equation is $\frac{1}{2}\log(xy) + \log 2 = \log(x+y)$.

Step 2: Combine the left side into one log.
$\frac{1}{2}\log(xy) + \log 2 = \log\left((xy)^{1/2}\right) + \log 2 = \log\left(2\sqrt{xy}\right)$. Setting this equal to $\log(x+y)$ gives $2\sqrt{xy} = x+y$, the same equation reached from a slightly different starting grouping.

Step 3: Recall the AM-GM inequality.
For positive numbers, the Arithmetic Mean $\frac{x+y}{2}$ is always greater than or equal to the Geometric Mean $\sqrt{xy}$, with equality only when $x = y$. Our equation says $\frac{x+y}{2} = \sqrt{xy}$ exactly, which is the equality case of AM-GM.

Step 4: Conclude from the equality condition.
AM equals GM only when both numbers being averaged are equal, so $x$ must equal $y$.

Step 5: Check with a number.
Try $x = y = 2$: left side is $\frac{1}{2}\log 2 + \frac{1}{2}\log 2 + \log 2 = \log 2 + \log 2 = \log 4$. Right side is $\log(2+2) = \log 4$. Both sides match, confirming x = y works.

Final Answer:
The relation holds only when x = y. \[ \boxed{x = y} \]