Solving the Exponential Equation for $2^k$

The problem asks us to find the value of $2^k$ given the equation:
\[ \left(\frac{7}{11}\right)^{k-5} = \left(\frac{11}{7}\right)^{k-9} \]

Step 1: Understanding the Equation Structure

We observe that the bases on both sides of the equation are related. The base on the right side, $\left(\frac{11}{7}\right)$, is the reciprocal of the base on the left side, $\left(\frac{7}{11}\right)$. We can use the property of exponents that states $a^{-n} = \frac{1}{a^n}$ or $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}$.

Step 2: Rewriting the Equation with a Common Base

Let's rewrite the right side of the equation using the base $\left(\frac{7}{11}\right)$. We know that $\left(\frac{11}{7}\right) = \left(\frac{7}{11}\right)^{-1}$. Substituting this into the equation:

\[ \left(\frac{7}{11}\right)^{k-5} = \left( \left(\frac{7}{11}\right)^{-1} \right)^{k-9} \]

Now, using the exponent rule $(a^m)^n = a^{m \times n}$, we simplify the right side:

\[ \left(\frac{7}{11}\right)^{k-5} = \left(\frac{7}{11}\right)^{-1 \times (k-9)} \]

\[ \left(\frac{7}{11}\right)^{k-5} = \left(\frac{7}{11}\right)^{-k+9} \]

Step 3: Equating the Exponents

Since the bases are now the same (and the base is not 0, 1, or -1), the exponents must be equal:

\[ k-5 = -k+9 \]

Step 4: Solving for $k$

We now solve this linear equation for $k$. First, add $k$ to both sides:

\[ k + k - 5 = 9 \] \[ 2k - 5 = 9 \]

Next, add 5 to both sides:

\[ 2k = 9 + 5 \] \[ 2k = 14 \]

Finally, divide by 2:

\[ k = \frac{14}{2} \] \[ k = 7 \]

Step 5: Calculating the Value of $2^k$

The question asks for the value of $2^k$. Now that we have found $k=7$, we substitute this value:

\[ 2^k = 2^7 \]

Let's calculate $2^7$:

$2^7 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 128$

Conclusion

Therefore, the value of $2^k$ is 128.