Find the solution of the equation
$\ln 3^{5x+1} = \ln 9^{x+5} + \log_2 16^{1-2x},$
Expressing your answer in terms of $\ln 3$. Solve $\log_4 (x-3) + \log_2 (2x+1)^{\frac{1}{2}} = 1$.

1. Simplify the logarithmic terms using the power law and change of base where necessary

$\ln 3^{5x+1} = (5x+1)\ln 3$
$\ln 9^{x+5} = \ln (3^2)^{x+5} = \ln 3^{2x+10} = (2x+10)\ln 3$
$\log_2 16^{1-2x} = \log_2 (2^4)^{1-2x} = \log_2 2^{4-8x} = 4-8x$

2. Substitute the simplified terms back into the equation and collect terms involving x

$(5x+1)\ln 3 = (2x+10)\ln 3 + 4 - 8x$
$5x\ln 3 + \ln 3 = 2x\ln 3 + 10\ln 3 + 4 - 8x$
$5x\ln 3 - 2x\ln 3 + 8x = 10\ln 3 - \ln 3 + 4$
$3x\ln 3 + 8x = 9\ln 3 + 4$

3. Factorise x and solve for x

$x(3\ln 3 + 8) = 9\ln 3 + 4$
$x = \frac{9\ln 3 + 4}{3\ln 3 + 8}$

1. Change the base of the first logarithmic term to base 2.

$\log_4 (x-3) = \frac{\log_2 (x-3)}{\log_2 4} = \frac{\log_2 (x-3)}{2} = \frac{1}{2} \log_2 (x-3)$

2. Simplify the second logarithmic term using the power law.

$\log_2 (2x+1)^{\frac{1}{2}} = \frac{1}{2} \log_2 (2x+1)$

3. Combine the logarithmic terms and solve for x.

$\frac{1}{2} \log_2 (x-3) + \frac{1}{2} \log_2 (2x+1) = 1$
$\log_2 (x-3) + \log_2 (2x+1) = 2$
$\log_2 [(x-3)(2x+1)] = 2$
$(x-3)(2x+1) = 2^2 = 4$
$2x^2 - 5x - 3 = 4$
$2x^2 - 5x - 7 = 0$
$(2x - 7)(x + 1) = 0$
$x = 3.5 \text{ or } x = -1$

4. Check for extraneous solutions.

For $x = -1$, $x-3 = -4 < 0$, so $\log_4(x-3)$ is undefined.
Therefore, $x = 3.5$ is the only valid solution.