If `lim_(n->oo)[((2009)^2010)^n+((2010)^2009)^n]^(1/n)` is equal to `a^b` where `a,b in N` then `a-b` is equal to:

Explanation

To solve the limit problem given by

\[ \lim_{n \to \infty} \left[ (2009^{2010})^n + (2010^{2009})^n \right]^{\frac{1}{n}}, \]

we will proceed step by step.

### Step 1: Rewrite the expression
We can rewrite the expression inside the limit as follows:

\[ \lim_{n \to \infty} \left[ (2009^{2010})^n + (2010^{2009})^n \right]^{\frac{1}{n}} = \lim_{n \to \infty} \left[ x^n + y^n \right]^{\frac{1}{n}}, \]

where we define \( x = 2009^{2010} \) and \( y = 2010^{2009} \).

### Step 2: Identify the dominant term
As \( n \) approaches infinity, the term with the larger base will dominate the sum. We need to compare \( x \) and \( y \):

\[ x = 2009^{2010}, \quad y = 2010^{2009}. \]

### Step 3: Compare \( x \) and \( y \)
To compare \( x \) and \( y \), we can take the ratio:

\[ \frac{y}{x} = \frac{2010^{2009}}{2009^{2010}} = \left(\frac{2010}{2009}\right)^{2009} \cdot \frac{1}{2009}. \]

Since \( \frac{2010}{2009} > 1 \), we can conclude that \( y > x \).

### Step 4: Simplify the limit
Since \( y \) is greater than \( x \), we can simplify the limit:

\[ \lim_{n \to \infty} \left[ x^n + y^n \right]^{\frac{1}{n}} = \lim_{n \to \infty} \left[ y^n \left( \left(\frac{x}{y}\right)^n + 1 \right) \right]^{\frac{1}{n}}. \]

### Step 5: Evaluate the limit
This can be simplified further:

\[ = \lim_{n \to \infty} y \left( \left(\frac{x}{y}\right)^n + 1 \right)^{\frac{1}{n}}. \]

As \( n \to \infty \), \( \left(\frac{x}{y}\right)^n \to 0 \) because \( \frac{x}{y} < 1 \). Thus, we have:

\[ \lim_{n \to \infty} \left( \left(\frac{x}{y}\right)^n + 1 \right)^{\frac{1}{n}} = 1. \]

So, the limit simplifies to:

\[ \lim_{n \to \infty} \left[ x^n + y^n \right]^{\frac{1}{n}} = y. \]

### Step 6: Substitute back for \( y \)
Now substituting back for \( y \):

\[ y = 2010^{2009}. \]

### Step 7: Express in the form \( a^b \)
We need to express \( 2010^{2009} \) in the form \( a^b \). Here, we can take:

\[ a = 2010, \quad b = 2009. \]

### Step 8: Calculate \( a - b \)
Finally, we need to find \( a - b \):

\[ a - b = 2010 - 2009 = 1. \]

Thus, the final answer is:

\[ \boxed{1}. \]