Question

If \(\log{\left(\frac{x+y}{3}\right)} = \frac{1}{2} \left(\log{x} + \log{y}\right)\) then the value of \(\frac{x}{y} + \frac{y}{x}\) is?

• A. 7
• B. 8
• C. -7
• D. 9

Hint:

To solve logarithmic equations, use properties such as \(\log{a} + \log{b} = \log{(ab)}\) and \(\frac{1}{2} \log{a} = \log{\sqrt{a}}\) to simplify the expressions.

Answer 1

The Correct Option is A

Given the equation: $\log\left(\frac{x+y}{3}\right) = \frac{1}{2}(\log x + \log y)$.
Using the logarithm property $\log a + \log b = \log(ab)$, we can rewrite the right side.
$\log\left(\frac{x+y}{3}\right) = \frac{1}{2}\log(xy)$.
Using the property $n \log a = \log(a^n)$, we get:
$\log\left(\frac{x+y}{3}\right) = \log\left((xy)^{\frac{1}{2}}\right)$.
$\log\left(\frac{x+y}{3}\right) = \log(\sqrt{xy})$.
Since the logarithms on both sides have the same base, their arguments must be equal.
$\frac{x+y}{3} = \sqrt{xy}$.
Squaring both sides of the equation:
$\left(\frac{x+y}{3}\right)^2 = (\sqrt{xy})^2$.
$\frac{(x+y)^2}{9} = xy$.
Expand the term $(x+y)^2$:
$\frac{x^2 + y^2 + 2xy}{9} = xy$.
Multiply both sides by 9:
$x^2 + y^2 + 2xy = 9xy$.
Subtract $2xy$ from both sides to isolate $x^2 + y^2$:
$x^2 + y^2 = 9xy - 2xy$.
$x^2 + y^2 = 7xy$.
To find the value of $\frac{x}{y} + \frac{y}{x}$, divide the entire equation by $xy$:
$\frac{x^2 + y^2}{xy} = \frac{7xy}{xy}$.
$\frac{x^2}{xy} + \frac{y^2}{xy} = 7$.
$\frac{x}{y} + \frac{y}{x} = 7$.