To solve the recurrence relation $T(n) = 9T\left(\frac{n}{6}\right) + n \log_2 n$ using the Master Method, we follow these steps:

Step 1: Identify Parameters for Master Method

The recurrence is in the form:

$T(n) = aT\left(\frac{n}{b}\right) + f(n)$

• Here, $a = 9$ (number of subproblems),
• $b = 6$ (factor by which the problem size is reduced),
• $f(n) = n \log_2 n$ (the non-recursive part).

Step 2: Calculate $n^{\log_b a}$

We need to find $n^{\log_b a}$ where:

$\log_b a = \log_6 9 = \frac{\log 9}{\log 6}$

Using the change of base formula with $\log_6 9$:

$\log_6 9 = \frac{\log 9}{\log 6} = \frac{\log (3^2)}{\log (2 \times 3)} = \frac{2 \log 3}{\log 2 + \log 3}$

Approximating $\log 3 \approx 0.4771$ and $\log 2 \approx 0.3010$:

$\log_6 9 \approx \frac{2 \times 0.4771}{0.3010 + 0.4771} = \frac{0.9542}{0.7781} \approx 1.226$

Thus, $n^{\log_6 9} = n^{1.226}$.

Step 3: Compare $f(n)$ with $n^{\log_b a}$

• $f(n) = n \log_2 n$
• $n^{\log_b a} = n^{1.226}$

To apply the Master Theorem, we compare $f(n)$ with $n^{\log_b a}$:

• $f(n) = O(n^{1.226 - \epsilon})$ for some $\epsilon > 0$ — Case 1
• $f(n) = \Theta(n^{1.226})$ — Case 2
• $f(n) = \Omega(n^{1.226 + \epsilon})$ and $af\left(\frac{n}{b}\right) \leq cf(n)$ for some $c < 1$ and sufficiently large $n$ — Case 3

Applying Case 2

• $f(n) = n \log_2 n$ grows slower than $n^{1.226}$, but they are not polynomially related because the $\log_2 n$ factor grows slower than any polynomial in $n$.
• However, for Case 2, $f(n) = \Theta(n^{\log_b a} \log^k n)$ is often a form considered when $k = 0$, and here $k = 1$. So, we need to re-check our basis, comparing against $n^{1.226}$.

Reassessing Case 2 with Specific Functions

Since $n \log_2 n = \Theta(n^{1.226})$, $f(n)$ is not precisely $\Theta(n^{1.226})$, suggesting we use a comparison:

• $f(n) = n \log_2 n$ is indeed $\Omega(n^{1.226 - \epsilon})$ for small $\epsilon$, indicating it is not quite a Case