Solve `log_4(log_3x)-log_(1//4)(log_(1//3)y)=0` and `x^2+y^2=17/4`.

Answer

To solve the equations \( \log_4(\log_3 x) - \log_{1/4}(\log_{1/3} y) = 0 \) and \( x^2 + y^2 = \frac{17}{4} \), we will follow these steps:

### Step 1: Simplify the logarithmic equation
We start with the equation:
\[ \log_4(\log_3 x) - \log_{1/4}(\log_{1/3} y) = 0 \]
This can be rewritten as:
\[ \log_4(\log_3 x) = \log_{1/4}(\log_{1/3} y) \]
Using the property of logarithms that states \( \log_{a}(b) = -\log_{1/a}(b) \), we can rewrite the right-hand side:
\[ \log_{1/4}(\log_{1/3} y) = -\log_4(\log_{1/3} y) \]
Thus, we have:
\[ \log_4(\log_3 x) = -\log_4(\log_{1/3} y) \]

### Step 2: Combine the logarithmic expressions
This implies:
\[ \log_4(\log_3 x) + \log_4(\log_{1/3} y) = 0 \]
Using the property of logarithms \( \log_a b + \log_a c = \log_a(bc) \), we can combine the logarithms:
\[ \log_4(\log_3 x \cdot \log_{1/3} y) = 0 \]
This means:
\[ \log_3 x \cdot \log_{1/3} y = 1 \]

### Step 3: Rewrite \( \log_{1/3} y \)
Using the change of base formula, we can express \( \log_{1/3} y \) as:
\[ \log_{1/3} y = \frac{1}{\log_3 y} \]
Substituting this back into our equation gives:
\[ \log_3 x \cdot \frac{1}{\log_3 y} = 1 \]
This simplifies to:
\[ \log_3 x = \log_3 y \]
Thus, we conclude:
\[ x = y \]

### Step 4: Substitute into the second equation
Now we substitute \( y = x \) into the second equation:
\[ x^2 + y^2 = \frac{17}{4} \]
This becomes:
\[ x^2 + x^2 = \frac{17}{4} \]
or:
\[ 2x^2 = \frac{17}{4} \]
Dividing both sides by 2 gives:
\[ x^2 = \frac{17}{8} \]

### Step 5: Solve for \( x \)
Taking the square root of both sides:
\[ x = \sqrt{\frac{17}{8}} = \frac{\sqrt{17}}{2\sqrt{2}} = \frac{\sqrt{34}}{4} \]
Since \( y = x \), we have:
\[ y = \frac{\sqrt{34}}{4} \]

### Step 6: Final values
The solutions for \( x \) and \( y \) are:
\[ x = \frac{\sqrt{34}}{4}, \quad y = \frac{\sqrt{34}}{4} \]