If `(a^(m))^(n)=a^(m^(n))`, then express 'm' in the terms of n is `(agt0, ane0, mgt1, ngt1)`

Options

A. `n^((1/(n-1)))`
B. `n^((1/(n+1)))`
C. `n^((1/n))`
D. None

Answer

To solve the equation \((a^{m})^{n} = a^{m^{n}}\) for \(m\) in terms of \(n\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ (a^{m})^{n} = a^{m^{n}} \]
Using the property of exponents \((x^{y})^{z} = x^{y \cdot z}\), we can rewrite the left side:
\[ a^{m \cdot n} = a^{m^{n}} \]

### Step 2: Set the exponents equal
Since the bases are the same (both are \(a\)), we can set the exponents equal to each other:
\[ m \cdot n = m^{n} \]

### Step 3: Rearrange the equation
We can rearrange this equation to isolate terms involving \(m\):
\[ m^{n} - m \cdot n = 0 \]

### Step 4: Factor the equation
We can factor out \(m\) from the left-hand side:
\[ m(m^{n-1} - n) = 0 \]

### Step 5: Analyze the factors
Since \(m > 1\) (as given in the problem), the first factor \(m = 0\) is not possible. Therefore, we must have:
\[ m^{n-1} - n = 0 \]

### Step 6: Solve for \(m\)
From the equation \(m^{n-1} = n\), we can express \(m\) in terms of \(n\):
\[ m = n^{\frac{1}{n-1}} \]

### Conclusion
Thus, the expression for \(m\) in terms of \(n\) is:
\[ m = n^{\frac{1}{n-1}} \]