If `((32)/(243))^(n)= (8)/(27)`, then find `((n+0.4)/(1024))^(-n)`

Answer

To solve the equation \(\left(\frac{32}{243}\right)^{n} = \frac{8}{27}\), we will follow these steps:

### Step 1: Rewrite the bases in terms of powers
We can express \(32\), \(243\), \(8\), and \(27\) as powers of their prime factors:
- \(32 = 2^5\)
- \(243 = 3^5\)
- \(8 = 2^3\)
- \(27 = 3^3\)

Thus, we can rewrite the equation:
\[ \left(\frac{2^5}{3^5}\right)^{n} = \frac{2^3}{3^3} \]

### Step 2: Apply the power of a quotient property
Using the property \(\left(\frac{a}{b}\right)^{m} = \frac{a^{m}}{b^{m}}\), we can simplify the left side:
\[ \frac{(2^5)^n}{(3^5)^n} = \frac{2^{5n}}{3^{5n}} \]

So the equation becomes:
\[ \frac{2^{5n}}{3^{5n}} = \frac{2^3}{3^3} \]

### Step 3: Set the numerators and denominators equal
Since the bases are the same, we can equate the exponents:
1. For the numerator:
\[ 5n = 3 \]
Solving for \(n\):
\[ n = \frac{3}{5} = 0.6 \]

### Step 4: Substitute \(n\) into \(\left(\frac{n + 0.4}{1024}\right)^{-n}\)
Now we need to calculate:
\[ \left(\frac{n + 0.4}{1024}\right)^{-n} \]
Substituting \(n = 0.6\):
\[ n + 0.4 = 0.6 + 0.4 = 1 \]
So we have:
\[ \left(\frac{1}{1024}\right)^{-0.6} \]

### Step 5: Rewrite \(1024\) as a power of \(2\)
We know that:
\[ 1024 = 2^{10} \]
Thus:
\[ \left(\frac{1}{2^{10}}\right)^{-0.6} = (2^{-10})^{-0.6} \]

### Step 6: Apply the power of a power property
Using the property \((a^m)^n = a^{m \cdot n}\):
\[ 2^{-10 \cdot (-0.6)} = 2^{6} = 64 \]

### Final Answer
Thus, the final value is:
\[ \boxed{64} \]