Set $\displaystyle y = 0$ and solve:

$\displaystyle y = \log _{2}(x+ 4)$

$\displaystyle \log _{2}(x+ 4) = 0$

$\displaystyle 2 ^ {\log _{2}(x+ 4) }= 2 ^ { 0}$

$\displaystyle x + 4 = 1$

$\displaystyle x + 4 - 4 = 1 - 4$

$\displaystyle x = -3$

The  $\displaystyle x$-intercept is $\displaystyle (-3,0)$.

Set $\displaystyle x= 0$ and evaluate $\displaystyle y$:

$\displaystyle y = \log _{3}(x+ 9)$

$\displaystyle y = \log _{3}(0+ 9) = \log _{3}9$

Since $\displaystyle 3 ^{2} =9$, 

$\displaystyle \log _{3}9 =2$, and the $\displaystyle y$-intercept is $\displaystyle (0,2)$.

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.