Question

If \(\frac 12 \ log\ x+\frac 12\ log\ y+log\ 2=log(x+y)\), then

• A. x = −y
• B. x = y + 1
• C. x = y
• D. y = x + 1

Answer 1

The Correct Option is C

To solve the given problem, we need to simplify the equation: \(\frac{1}{2} \log x + \frac{1}{2} \log y + \log 2 = \log(x + y)\). We will use the properties of logarithms to simplify both sides.
Step 1: Combine the logarithmic terms on the left side using the properties of logarithms. The formula \(\log a + \log b = \log(ab)\) helps us combine logs: 
 

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

Step 2: Use the rule of logarithms, \(a \log b = \log(b^a)\):
 

\(\frac{1}{2} \log(xy) = \log((xy)^{1/2}) = \log\sqrt{xy}\)

Step 3: Combine with the remaining logarithmic term using the addition rule:
 

\(\log\sqrt{xy} + \log 2 = \log(2\sqrt{xy})\)

Step 4: Equating the simplified expression to the right side of the original equation:
 

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

Since the equation \(\log a = \log b\) implies \(a = b\), we get:
 

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

Step 5: Square both sides to remove the square root:
 

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

\(\Rightarrow 4xy = x^2 + 2xy + y^2\)
Step 6: Rearrange the terms to form a quadratic equation:
 

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

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

Step 7: The equation \((x - y)^2 = 0\) implies that \(x - y = 0\):
 

\(x = y\)

Hence, the correct solution is \(x = y\), which corresponds to the correct answer from the given options.