Solve $9^{x+1} - 3^{x+2} = 2\left(3^{x+1} - 2\right)$. Given that $\log_3(2x-1) = \log_9(2x-1) + 1$, find the value of $x$. Given that $\log_a xy = m$ and $\log_a \frac{x^3}{y^2} = n$, find $5\log_a \left(\frac{x}{y}\right)$ in terms of $m$ and $n$.

1. Express all exponential terms with a common base of 3

$9^{x+1} - 3^{x+2} = 2\left(3^{x+1} - 2\right)$
$(3^2)^{x+1} - 3^2 \cdot 3^x = 2\left(3 \cdot 3^x - 2\right)$
$3^{2x+2} - 9(3^x) = 6(3^x) - 4$
$9(3^x)^2 - 9(3^x) = 6(3^x) - 4$

2. Use substitution to form a quadratic equation

Let $u = 3^x$.
$9u^2 - 9u = 6u - 4$
$9u^2 - 15u + 4 = 0$

3. Solve the quadratic equation for $u$

$(3u - 1)(3u - 4) = 0$
$u = \frac{1}{3} \quad \text{or} \quad u = \frac{4}{3}$

4. Solve for $x$ by substituting back $3^x = u$

When $u = \frac{1}{3}$:
$3^x = 3^{-1} \implies x = -1$

When $u = \frac{4}{3}$:
$3^x = \frac{4}{3}$
$x \ln 3 = \ln\left(\frac{4}{3}\right)$
$x = \frac{\ln(4/3)}{\ln 3} \approx 0.26185 \approx 0.262 \text{ (to 3 s.f.)}$

1. Change the base of the logarithm from 9 to 3

$\log_9(2x-1) = \frac{\log_3(2x-1)}{\log_3 9} = \frac{\log_3(2x-1)}{2}$

2. Substitute back into the equation and simplify

$\log_3(2x-1) = \frac{1}{2}\log_3(2x-1) + 1$
$\log_3(2x-1) - \frac{1}{2}\log_3(2x-1) = 1$
$\frac{1}{2}\log_3(2x-1) = 1$
$\log_3(2x-1) = 2$

3. Convert to exponential form and solve for $x$

$2x - 1 = 3^2$
$2x - 1 = 9$
$2x = 10$
$x = 5$

1. Expand the given logarithmic equations using the product and quotient laws of logarithms

$\log_a x + \log_a y = m$ --- (1)
$3\log_a x - 2\log_a y = n$ --- (2)

2. Solve the simultaneous equations for $\log_a x$ and $\log_a y$

From (1), $\log_a y = m - \log_a x$.
Substitute into (2):
$3\log_a x - 2(m - \log_a x) = n$
$5\log_a x - 2m = n$
$5\log_a x = 2m + n$
$\log_a x = \frac{2m + n}{5}$

Substitute back into (1):
$\frac{2m + n}{5} + \log_a y = m$
$\log_a y = m - \frac{2m + n}{5} = \frac{3m - n}{5}$

3. Evaluate the target expression $5\log_a \left(\frac{x}{y}\right)$

$5\log_a \left(\frac{x}{y}\right) = 5(\log_a x - \log_a y)$
$= 5\left( \frac{2m + n}{5} - \frac{3m - n}{5} \right)$
$= (2m + n) - (3m - n)$
$= 2n - m$