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 property. 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 the following relationship:

\[  \log_ab = \frac{\log_cb}{\log_ca}  \]

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

\[  \log_24 + \log_2(23-x) = \frac{\log_2(x-8)}{\log_2\sqrt{2}}   \]

\[  \log_24(23-x) = \frac{\log_2(x-8)}{\log_22^{1/2}}       \]

\[  \log_2(92-4x) = \frac{\log_2(x-8)}{\frac{1}{2}}       \]

\[  \log_2(92-4x) = 2\log_2(x-8)            \]

\[  \log_2(92-4x) = \log_2(x-8)^2            \]

\[  92-4x = (x-8)^2            \]

\[  92-4x = x^2-16x+64         \]

\[  x^2-12x-28 = 0           \]

\[  (x-14)(x+2) = 0   \]

We can have \(x=14\) or \(x=-2\). However, the domain must be \(8<x<23\), so the intersection is \(x=14\).