The graph of a logarithmic function has a vertical asymptote which can be found by finding the value at which the power is equal to 0:

\(\displaystyle x-7 = 0 \Rightarrow x = 7\)

If \(\displaystyle x \leq 7\), then \(\displaystyle \log_{5} (x-7)\) is an undefined expression, so the vertical asymptote is \(\displaystyle x = 7\).

Set \(\displaystyle g(x) =0\) and evaluate \(\displaystyle x\) to find the \(\displaystyle x\)-coordinate of the \(\displaystyle x\)-intercept.

\(\displaystyle g(x) =2 \log_{4}(x-4)+ 3 = 0\)

\(\displaystyle 2 \log_{4}(a-4) = -3\)

\(\displaystyle \log_{4}(a-4) = -\frac{3}{2}\)

This can be rewritten in exponential form:

\(\displaystyle a - 4 = 4 ^{-\frac{3}{2}}= \frac{1}{4 ^{ \frac{3}{2}}} = \frac{1}{(\sqrt{4 })^{ 3}} = \frac{1}{2^{ 3}} = \frac{1}{8}\)

\(\displaystyle a = 4\frac{1}{8}\)

The \(\displaystyle x\)-intercept of the graph of \(\displaystyle g\) is \(\displaystyle \left ( 4\frac{1}{8}, 0 \right )\).

The \(\displaystyle y\:\)-coordinate of the \(\displaystyle y\:\)-intercept is \(\displaystyle g(0)\):

\(\displaystyle g(x) = \log_{9}(x-4)+ 2\)

\(\displaystyle g(0) = \log_{9}(0-4)+ 2\)

\(\displaystyle = \log_{9}( -4)+ 2\)

However, the logarithm of a negative number is an undefined expression, so \(\displaystyle g(0)\) is an undefined quantity, and the graph of \(\displaystyle g\) has no \(\displaystyle y\:\)-intercept.

Only positive numbers have logarithms, so

\(\displaystyle x+2 > 0\)

\(\displaystyle x> -2\)

The graph never crosses the vertical line of the equation \(\displaystyle x = -2\), so this is the vertical asymptote. 

Only positive numbers have logarithms, so

\(\displaystyle x- 4 > 0\)

\(\displaystyle x> 4\)

The graph never crosses the vertical line of the equation \(\displaystyle x = 4\), so this is the vertical asymptote. 

The \(\displaystyle y\:\)-coordinate of the \(\displaystyle y\:\)-intercept is \(\displaystyle g(0)\):

\(\displaystyle g(x) =-2 \log_{8}(x+2)+ 3\)

\(\displaystyle g(0) =-2 \log_{8}(0 +2)+ 3\)

\(\displaystyle g(0) =-2 \log_{8}2+ 3\)

Since 2 is the cube root of 8, \(\displaystyle 8^{\frac{1}{3}}= 2\), and  \(\displaystyle \log_{8}2 = \frac{1}{3}\). Therefore, 

\(\displaystyle g(0) =-2 \cdot \frac{1}{3}+ 3 = -\frac{2}{3}+ 3 = 2\frac{1}{3}\).

The \(\displaystyle y\:\)-intercept is \(\displaystyle \left ( 2\frac{1}{3},0 \right )\).

Since \(\displaystyle - \log N = \log \left ( \frac{1}{N} \right )\), the definition of \(\displaystyle f\) can be rewritten as follows:

\(\displaystyle f(x) = - \log (x+3)\)

First, we need to find the \(\displaystyle x\)-coordinate of the point at which the graphs of \(\displaystyle f\) and \(\displaystyle g\) meet by setting 

\(\displaystyle g(x)= f(x)\)

\(\displaystyle \log (4x+12)= \log \frac{1}{x+3}\)

Since the common logarithms of the polynomial and the rational expression are equal, we can set those expressions themselves equal, then solve:

\(\displaystyle (4x+12)= \frac{1}{x+3}\)

\(\displaystyle (4x+12)(x+3)= \frac{1}{x+3} \cdot (x+3)\)

\(\displaystyle 4x^{3}+24x+36 =1\)

\(\displaystyle 4x^{3}+24x+35 =0\)

We can solve using the \(\displaystyle ac\) method, finding two integers whose sum is 24 and whose product is \(\displaystyle 4 \cdot 35 = 140\) - these integers are 10 and 14, so we split the niddle term, group, and factor: 

\(\displaystyle (4x^{3}+10x)+(14x+35 )=0\)

\(\displaystyle 2x (2x+5)+7(2x+5 )=0\)

\(\displaystyle (2x +7)(2x+5 )=0\)

\(\displaystyle 2x+7 = 0\)

\(\displaystyle 2x= -7\)

\(\displaystyle x = -\frac{7}{2} = -3\frac{1}{2}\)

or 

\(\displaystyle 2x+5= 0\)

\(\displaystyle 2x= -5\)

\(\displaystyle x = -\frac{5}{2} = -2\frac{1}{2}\)

This gives us two possible \(\displaystyle x\)-coordinates. However, since 

\(\displaystyle g\left ( -3\frac{1}{2} \right ) = \log \left [4\left ( -3\frac{1}{2} \right ) +12 \right ] = \log (-14+12) = \log (-2)\),

an undefined quantity - negative numbers not having logarithms -

we throw this value out. As for the other \(\displaystyle x\)-value, we evaluate: 

\(\displaystyle f\left ( -2\frac{1}{2} \right ) = \log \frac{1}{ -2\frac{1}{2} +3} = \log \frac{1}{\frac{1}{2}} = \log 2\)

and 

\(\displaystyle g\left ( -2\frac{1}{2} \right ) = \log \left [4\left ( -2\frac{1}{2} \right ) +12 \right ] = \log (-10+12) = \log 2\)

\(\displaystyle -2 \frac{1}{2}\) is the correct \(\displaystyle x\)-value, and \(\displaystyle \log 2\) is the correct \(\displaystyle y\)-value.

Substitute \(\displaystyle u\) and \(\displaystyle v\) for \(\displaystyle \log_{2} x\) and \(\displaystyle \log_{3} y\), respectively, and solve the resulting system of linear equations:

\(\displaystyle 2 u-3 v = 13\)

\(\displaystyle 5 u+2 v = 4\)

Multiply the first equation by 2, and the second by 3, on both sides, then add:

\(\displaystyle 4 u-6 v = 26\)

\(\displaystyle \underline{15 u+6 v = 12}\)

\(\displaystyle 19u\)            \(\displaystyle =38\)

\(\displaystyle u = 38 \div 19 = 2\)

Now back-solve:

\(\displaystyle 5 (2)+2 v = 4\)

\(\displaystyle 10+2 v = 4\)

\(\displaystyle 2v=-6\)

\(\displaystyle v= -3\)

We need to find both \(\displaystyle x\) and \(\displaystyle y\) to ensure a solution exists. By substituting back:

\(\displaystyle u = 2\)

\(\displaystyle \log_{2} x = 2\)

\(\displaystyle x = 2^{2} = 4\).

\(\displaystyle v= -3\)

\(\displaystyle \log_{3} y = -3\)

\(\displaystyle y = 3^{-3} = \frac{1}{3^{3}} = \frac{1}{27}\)

\(\displaystyle \left (4, \frac{1}{27} \right )\) is the solution, and \(\displaystyle y = \frac{1}{27}\), the correct choice.

Substitute \(\displaystyle u\) and \(\displaystyle v\) for \(\displaystyle \log x\) and \(\displaystyle \log y\), respectively, and solve the resulting system of linear equations:

\(\displaystyle 2 u+3 v = 12\)

\(\displaystyle 5 u- 2 v = 11\)

Multiply the first equation by 2, and the second by 3, on both sides, then add:

\(\displaystyle 4 u+6 v = 24\)

\(\displaystyle \underline{15 u- 6 v = 33}\)

\(\displaystyle 19u\)            \(\displaystyle =57\)

\(\displaystyle u = 57 \div 19 = 3\)

Back-solve:

\(\displaystyle 2 (3)+3 v = 12\)

\(\displaystyle 6+3 v = 12\)

\(\displaystyle 3v = 6\)

\(\displaystyle v = 2\)

We need to find both \(\displaystyle x\) and \(\displaystyle y\) to ensure a solution exists. By substituting back:

\(\displaystyle u = 3\)

\(\displaystyle \log x = 3\)

\(\displaystyle x=10^{3} = 1,000\)

and

\(\displaystyle v=2\)

\(\displaystyle \log y = 2\)

\(\displaystyle y=10^{2} = 100\)

We check this solution in both equations:

\(\displaystyle 2 \log x+3 \log y = 12\)

\(\displaystyle 2 \log 1,000+3 \log 100 = 12\)

\(\displaystyle 2 (3)+3 (2)= 12\)

\(\displaystyle 6+6 = 12\) - true.

\(\displaystyle 5 \log x- 2 \log y = 11\)

\(\displaystyle 5 \log 1,000- 2 \log 100 = 11\)

\(\displaystyle 5 (3) -2 (2) = 11\)

\(\displaystyle 15-4 = 11\) - true.

\(\displaystyle (1000, 100)\) is the solution, and \(\displaystyle x=1,000\), the correct choice.

Given the function \(\displaystyle f(x) = A \log (x-B)+C\), the vertical asymptote can be found by observing that a logarithm cannot be taken of a number that is not positive. Therefore, it must hold that \(\displaystyle x-B >0\), or, equivalently, \(\displaystyle x>B\) and that the graph of \(\displaystyle f\) will never cross the vertical line \(\displaystyle x=B\). That makes \(\displaystyle x=B\) the vertical asymptote, so it follows that the graph with vertical asymptote \(\displaystyle x = 5\) will have \(\displaystyle 5\) in the \(\displaystyle B\) position. The only choice that meets this criterion is

\(\displaystyle f(x) = 3 \log (x-5)+7\)