The equation $\log_2 x + \log_8 x = \log_7 49$ has the solution $x = 2^a$.
Find the value of $a$. Show that the equation $\log_3(4x - 11) - \log_3(x - 3) = 1$ has no solution.

1. Evaluate the constant term on the right hand side

$\log_7 49 = \log_7 (7^2) = 2$

2. Change the base of $\log_8 x$ to base 2

$\log_8 x = \frac{\log_2 x}{\log_2 8} = \frac{\log_2 x}{3}$

3. Substitute back into the equation and solve for $\log_2 x$

$\log_2 x + \frac{1}{3}\log_2 x = 2$

$\frac{4}{3}\log_2 x = 2$

$\log_2 x = 2 \times \frac{3}{4} = \frac{3}{2}$

4. Convert from logarithmic to exponential form to find $a$

$x = 2^{\frac{3}{2}}$

Since $x = 2^a$, $a = \frac{3}{2}$

1. Combine the logarithmic terms using the quotient law

$\log_3 \left( \frac{4x - 11}{x - 3} \right) = 1$

2. Convert the logarithmic equation to exponential form

$\frac{4x - 11}{x - 3} = 3^1$

3. Solve the resulting linear equation for $x$

$4x - 11 = 3(x - 3)$
$4x - 11 = 3x - 9$
$x = 2$

4. Check the validity of the solution against the domain of the original logarithms

Substitute $x = 2$ into the original arguments:
$4(2) - 11 = -3$
$2 - 3 = -1$
Since the argument of a logarithm must be strictly positive ($>0$), $\log_3(-3)$ and $\log_3(-1)$ are undefined.
Therefore, $x = 2$ is an extraneous solution, and the equation has no valid solution.