Equations of logarithmic functions

Question:
There are two functions:
$y = 2+\log_2(23-x) $
$y = \log_{\sqrt{2}}(x-8)$
Find the $x$-coordinate of the intersection of these curves.

Answer:
A general logarithmic function reads
\[
y = \log(X)
\]
Within real numbers, $X$ must be positive. The functions given above can also be found the domains in terms of this feature. From the function, $y = 2+\log_2(23-x)$, the domain must be
\[
23-x>0   \\
\rightarrow x < 23
\]
Likewise, the second function has the domain:
\[
x>8
\]
Thus, we can estimate the position of the intersection should be $8<x<23$. (See the figure.)
Now, equate both functions:
\[
2+\log_2(23-x) = \log_{\sqrt{2}}(x-8)
\]
Let us use
\[
\log_ab = \frac{\log_cb}{\log_ca}
\]
The base of right hand side can be changed into 2, and the constant in left hand side, 2, can become $\log_24$. Then, we have
\begin{eqnarray*}
\log_24 + \log_2(23-x) &=& \frac{\log_2(x-8)}{\log_2\sqrt{2}}   \\
\log_24(23-x) &=& \frac{\log_2(x-8)}{\log_22^{1/2}}       \\
\log_2(92-4x) &=& \frac{\log_2(x-8)}{\frac{1}{2}}       \\
\log_2(92-4x) &=& 2\log_2(x-8)            \\
\log_2(92-4x) &=& \log_2(x-8)^2            \\
92-4x &=& (x-8)^2            \\
92-4x &=& x^2-16x+64         \\
x^2-12x-28 &=& 0           \\
(x-14)(x+2) &=& 0
\end{eqnarray*}
We can have $x=14$ or $x=-2$. However, the domain must be $8<x<23$, so the intersection is $x=14$.

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