Solve this function for me: g (x) = a log4 (b(x - h)) + k . It passes through the points (x,y) (5, 4.5) and (13, 3.5)

We are given the function g(x) = a\log_{4}(b(x-h)) + k and two points that the function passes through: (5, 4.5) and (13, 3.5).

Using the point (5, 4.5), we get:
4.5 = a\log_{4}(b(5-h)) + k

And using the point (13, 3.5), we get:
3.5 = a\log_{4}(b(13-h)) + k

Now, we have a system of two equations:
\begin{cases}\begin{cases}\end{cases}

Subtracting the second equation from the first:
4.5 - 3.5 = a(\log_{4}(5h) - \log_{4}(13h))
1 = a(\log_{4}(\frac{5h}{13h}))
1 = \log_{4}(\frac{5}{13})a
a = \frac{1}{\log_{4}(\frac{5}{13})}

Then, substitute back to solve for k :
4.5 = \frac{1}{\log_{4}(\frac{5}{13})} \log_{4}(5h) + k
4.5 = \frac{\log_{4}(5h)}{\log_{4}(\frac{5}{13})} + k
4.5 = \log_{\frac{5}{13}}(5h) + k
4.5 = \frac{\ln(5h)}{\ln(\frac{5}{13})} + k
4.5 = \frac{\ln(5) + \ln(h)}{\ln(\frac{5}{13})} + k
4.5 = \frac{\ln(5) + \ln(h)}{\ln(5) - \ln(13)} + k
4.5 = \frac{\ln(5h)}{\ln(5) - \ln(13)} + k
4.5 = \log_{5/13}(5h) + k
4.5 = \log_{5/13}(5) + \log_{5/13}(h) + k
4.5 = 1 + \log_{5/13}(h) + k
3.5 = \log_{5/13}(h) + k
3.5 -1 = \log_{5/13}(h)
2.5 = \log_{5/13}(h)
h = \left( \frac{5}{13} \right)^{2.5}

Therefore, the values of the constants a , b , h , and k are:
a = \frac{1}{\log_{4}\left(\frac{5}{13} \right)}
h = \left(\frac{5}{13}\right)^{2.5}

\textbf{Answer:} The values of the constants a and h are a = \frac{1}{\log_{4}\left(\frac{5}{13} \right)} and h = \left(\frac{5}{13}\right)^{2.5} .