Use the substitution $u = 7^x$, solve the equation $7^{2x+2}-4(7^{x+1})-5=0$ and show that the solution may be written in the form $\log_a b -1$ where $a$ and $b$ are integers to be determined. Solve the equation $\log_3(2x-1)-\log_9(x^2+2)=\log_{25}5$.

1. Express the equation in terms of $u = 7^x$.

$7^2(7^x)^2 - 4(7)(7^x) - 5 = 0$
$49u^2 - 28u - 5 = 0$

2. Solve the quadratic equation for $u$.

$(7u - 5)(7u + 1) = 0$
$u = \frac{5}{7} \quad \text{or} \quad u = -\frac{1}{7}$

3. Substitute back $u = 7^x$ and reject the invalid solution.

$7^x = \frac{5}{7} \quad \text{or} \quad 7^x = -\frac{1}{7}$
Since $7^x > 0$ for all real $x$, $7^x = -\frac{1}{7}$ has no real solution.

4. Solve for $x$ and express it in the required form.

$x = \log_7\left(\frac{5}{7}\right)$
$x = \log_7 5 - \log_7 7$
$x = \log_7 5 - 1$
Comparing with $\log_a b - 1$, we have $a = 7$ and $b = 5$.

1. Evaluate the constant term on the right hand side.

$\log_{25}5 = \frac{1}{2}$

2. Change the base of the logarithm to base 3.

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

3. Combine the logarithmic terms into a single logarithm.

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

4. Convert the logarithmic equation to an algebraic equation and solve for $x$.

$\frac{(2x-1)^2}{x^2+2} = 3^1 = 3$
$4x^2 - 4x + 1 = 3x^2 + 6$
$x^2 - 4x - 5 = 0$
$(x-5)(x+1) = 0$
$x = 5 \text{ or } x = -1$

5. Check for extraneous solutions.

When $x = -1$, $2x-1 = -3 < 0$, so $\log_3(2x-1)$ is undefined. Thus, $x = -1$ is rejected.
When $x = 5$, $2x-1 = 9 > 0$ and $x^2+2 = 27 > 0$, which are valid.
Therefore, $x = 5$.