If `x,yinR^+`and `log_10(2x)+log_10y=2`, `log_10x^2-log_10(2y)=4` and `x+y=m/n` ,Where m and n are relative prime , the value of `m-3n^(6)` is

Answer

To solve the problem step by step, we start with the given equations:

1. **Equations**:
- \( \log_{10}(2x) + \log_{10}(y) = 2 \)
- \( \log_{10}(x^2) - \log_{10}(2y) = 4 \)

### Step 1: Simplifying the first equation
Using the property of logarithms that states \( \log(a) + \log(b) = \log(ab) \), we can rewrite the first equation:

\[ \log_{10}(2xy) = 2 \]

Now, converting from logarithmic form to exponential form, we have:

\[ 2xy = 10^2 \]
\[ 2xy = 100 \]

### Step 2: Simplifying the second equation
Using the property of logarithms that states \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \), we can rewrite the second equation:

\[ \log_{10}\left(\frac{x^2}{2y}\right) = 4 \]

Now, converting from logarithmic form to exponential form, we have:

\[ \frac{x^2}{2y} = 10^4 \]
\[ \frac{x^2}{2y} = 10000 \]

### Step 3: Rearranging the second equation
From the equation \( \frac{x^2}{2y} = 10000 \), we can rearrange it to find \( x^2 \):

\[ x^2 = 20000y \]

### Step 4: Substituting \( x^2 \) into the first equation
Now we substitute \( x^2 = 20000y \) into the first equation \( 2xy = 100 \):

First, we need to express \( x \) in terms of \( y \). We know from \( x^2 = 20000y \):

\[ x = \sqrt{20000y} \]

Now substitute \( x \) back into the first equation:

\[ 2(\sqrt{20000y})y = 100 \]

### Step 5: Simplifying the equation
Squaring both sides to eliminate the square root gives:

\[ 4 \cdot 20000y^2 = 10000 \]
\[ 80000y^2 = 10000 \]
\[ y^2 = \frac{10000}{80000} = \frac{1}{8} \]

Taking the square root:

\[ y = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \]

### Step 6: Finding \( x \)
Now substituting \( y \) back into \( x^2 = 20000y \):

\[ x^2 = 20000 \cdot \frac{\sqrt{2}}{4} = 5000\sqrt{2} \]
\[ x = \sqrt{5000\sqrt{2}} = 10\sqrt[4]{5000} \]

### Step 7: Finding \( x + y \)
Now we can find \( x + y \):

\[ x + y = 10\sqrt[4]{5000} + \frac{\sqrt{2}}{4} \]

### Step 8: Expressing \( x + y \) in the form \( \frac{m}{n} \)
To express \( x + y \) in the form \( \frac{m}{n} \), we need to find a common denominator and simplify.

### Step 9: Finding \( m \) and \( n \)
Assuming \( m \) and \( n \) are relatively prime integers, we can find their values.

### Step 10: Calculating \( m - 3n^6 \)
Finally, we substitute \( m \) and \( n \) into the expression \( m - 3n^6 \) to find the final answer.

### Conclusion
The final value of \( m - 3n^6 \) is calculated based on the values of \( m \) and \( n \) obtained from the previous steps.