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

To solve the equation \(\frac{1}{2}\log x + \frac{1}{2}\log y + \log 2 = \log(x+y)\), we will use the properties of logarithms. Specifically, we'll use the following properties:

• \(\log a + \log b = \log (ab)\) 
• \(c \log a = \log (a^c)\)

Given the equation:

\(\frac{1}{2}\log x + \frac{1}{2}\log y + \log 2 = \log(x+y)\)

First, we express the terms with coefficients as powers:

\(\log x^{1/2} + \log y^{1/2} + \log 2 = \log(x+y)\)

Using the first property of logarithms, combine them:

\(\log((xy)^{1/2} \cdot 2) = \log(x+y)\)

Since the logarithms are equal, their arguments must be equal:

\((xy)^{1/2} \cdot 2 = x+y\)

Simplifying further, we have:

\(2 \sqrt{xy} = x + y\)

Squaring both sides to remove the square root:

\(4xy = (x+y)^2\)

Expanding the right side:

\(4xy = x^2 + 2xy + y^2\)

Rearranging the terms gives:

\(x^2 + 2xy + y^2 - 4xy = 0\)

Which simplifies to:

\(x^2 - 2xy + y^2 = 0\)

This expression can be factored as:

\((x-y)^2 = 0\)

The principal root of this equation is:

\(x = y\)

Thus, the correct answer is \(x = y\).

Answer 2

Step 1: Combine the log terms on the left using log rules.
The rule \(m\log a = \log(a^m)\) turns \(\frac{1}{2}\log x\) into \(\log(x^{1/2})\) and \(\frac{1}{2}\log y\) into \(\log(y^{1/2})\). So the left side becomes \(\log(x^{1/2}) + \log(y^{1/2}) + \log 2\).

Step 2: Merge everything into a single log using the product rule.
Since \(\log A + \log B = \log(AB)\), the left side is \(\log(2\sqrt{x}\sqrt{y}) = \log(2\sqrt{xy})\). The equation now reads \(\log(2\sqrt{xy}) = \log(x+y)\).

Step 3: Remove the log from both sides.
Log is a one to one function, so if \(\log P = \log Q\) then \(P = Q\). This gives \(2\sqrt{xy} = x + y\).

Step 4: Rearrange into a perfect square.
Move everything to one side: \(x + y - 2\sqrt{xy} = 0\). Since \(x = (\sqrt{x})^2\) and \(y = (\sqrt{y})^2\), this is \((\sqrt{x})^2 + (\sqrt{y})^2 - 2\sqrt{x}\sqrt{y} = 0\), which is exactly the expansion of \((\sqrt{x} - \sqrt{y})^2\).

Step 5: Solve the squared term.
A square can only be zero if the term inside is zero, so \(\sqrt{x} - \sqrt{y} = 0\), which gives \(\sqrt{x} = \sqrt{y}\) and so \(x = y\).

Final Answer:
The condition simplifies to x = y. \[ \boxed{x = y} \]