If `log_(2)x+log_(x)2=(10)/(3)=log_(2)y+log_(y)2` and `xney` then `x+y` =

Options

A. `2`
B. `65//8`
C. `37//6`
D. none of these

Answer

To solve the equation \( \log_{2}x + \log_{x}2 = \frac{10}{3} = \log_{2}y + \log_{y}2 \) and find \( x + y \) given that \( x \neq y \), we can follow these steps:

### Step 1: Rewrite the logarithmic expressions
Using the change of base formula, we can rewrite \( \log_{x}2 \) as \( \frac{1}{\log_{2}x} \). Thus, we can rewrite the equation as:
\[ \log_{2}x + \frac{1}{\log_{2}x} = \frac{10}{3} \]
Let \( t = \log_{2}x \). Then, we have:
\[ t + \frac{1}{t} = \frac{10}{3} \]

### Step 2: Multiply through by \( t \)
To eliminate the fraction, multiply both sides by \( t \):
\[ t^2 + 1 = \frac{10}{3}t \]
Rearranging gives us:
\[ 3t^2 - 10t + 3 = 0 \]

### Step 3: Solve the quadratic equation
We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3, b = -10, c = 3 \):
\[ t = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
Calculating the discriminant:
\[ t = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} \]
This gives us two solutions:
\[ t_1 = \frac{18}{6} = 3 \quad \text{and} \quad t_2 = \frac{2}{6} = \frac{1}{3} \]

### Step 4: Find \( x \) and \( y \)
Since \( t = \log_{2}x \), we can find \( x \):
1. If \( t = 3 \), then \( x = 2^3 = 8 \).
2. If \( t = \frac{1}{3} \), then \( x = 2^{1/3} = \sqrt[3]{2} \).

Now, we have two pairs of values for \( x \) and \( y \):
- If \( x = 8 \), then \( \log_{2}y + \log_{y}2 = \frac{10}{3} \) will yield \( y = 2 \) (since \( \log_{2}2 = 1 \)).
- If \( x = \sqrt[3]{2} \), then \( y = 4 \) (since \( \log_{2}4 = 2 \)).

### Step 5: Calculate \( x + y \)
Now we can calculate \( x + y \):
1. If \( x = 8 \) and \( y = 2 \):
\[ x + y = 8 + 2 = 10 \]
2. If \( x = \sqrt[3]{2} \) and \( y = 4 \):
\[ x + y = \sqrt[3]{2} + 4 \]
However, since \( x \) and \( y \) must be distinct and we already have \( x = 8 \) and \( y = 2 \) as a valid pair, we take \( x + y = 10 \).

Thus, the final answer is:
\[ \boxed{10} \]