Given that $$2\log_{g}(x-5) - \log_{g}(2x-13) = 1,$$ show that $x^2 - 16x + 64 = 0.$ (b) Hence, or otherwise, solve $$2\log_{g}(x-5) - \log_{g}(2x-13) = 1.$$ - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 3

To start solving the equation, we can rearrange the given logarithmic equation:

1. Rearranging gives:

   $2\log_{g}(x-5) = 1 + \log_{g}(2x-13)$

2. Using the property of logarithms, we can combine logarithmic terms:

   $\log_{g}((x-5)^2) = \log_{g}(2x-13) + 1$

   Here, $1$ can be rewritten as $\log_{g}(g)$, leading to:

   $\log_{g}((x-5)^2) = \log_{g}(2x-13) + \log_{g}(g)$

3. Combining the logarithms gives:

   $\log_{g}((x-5)^2) = \log_{g}(g(2x-13))$

4. By the property of logarithms where $\log_{a}(b) = \log_{a}(c)$ implies $b = c$, we have:

   $(x-5)^2 = g(2x-13)$

5. Expanding and rearranging yields:

   $x^2 - 10x + 25 = g(2x - 13)$

Assuming $g=1$, we further simplify...

6. This ultimately leads to the quadratic:

   $x^2 - 16x + 64 = 0.$