Let \(\displaystyle g(x) = \ln x\)

 In terms of \(\displaystyle g(x)\),

\(\displaystyle f(x) =3 g (x+ 4)+ 2\)

This is the graph of \(\displaystyle g(x)\) shifted left 4 units, stretched vertically by a factor of 3, then shifted up 2 units. 

The graph of \(\displaystyle g(x)\) does not have a horizontal asymptote; therefore, a transformation of this graph, such as that of \(\displaystyle f(x)\), does not have a horizontal asymptote either.

Let \(\displaystyle g(x) = \ln x\). In terms of \(\displaystyle g(x)\),

\(\displaystyle f(x) =3 g (x+ 4)+ 2\).

The graph of \(\displaystyle g(x)\) has as its vertical asymptote the line of the equation \(\displaystyle x= 0\). The graph of \(\displaystyle f(x)\) is the result of three transformations on the graph of \(\displaystyle g(x)\)- a left shift of 4 units \(\displaystyle (+4)\), a vertical stretch ( \(\displaystyle 3g\) ), and an upward shift of 2 units ( \(\displaystyle +2\) ). Of the three transformations, only the left shift affects the position of the vertical asymptote - the asymptote of \(\displaystyle f(x)\) also shifts left 4 units, to \(\displaystyle x = -4\).

\(\displaystyle \left (\frac{f^{-3}}{g^{-2}} \right )^{4}=\left (\frac{g^{2}}{f^{3}} \right )^{4}=\frac{g^{2\times 4}}{f^{3\times4}}=\frac{g^{8}}{f^{12}}\)

\(\displaystyle \left (\frac{34x}{2y} \right )(17xy)^{-1}=\left (\frac{34x}{2y} \right )\left (\frac{1}{17xy} \right )=\frac{34x}{34xy^2}=\frac{1}{y^2}\)

\(\displaystyle \left (\frac{m}{n} \right )^{-2}\left (\frac{n}{m} \right )^{4}=\left (\frac{n}{m} \right )^{2}\left (\frac{n}{m} \right )^{4}=\frac{n^{2+4}}{m^{2+4}}=\frac{n^{6}}{m^{6}}\)

\(\displaystyle \frac{(a^{2})^{-7}b^{0}}{a^{3}b^{-4}}=\frac{a^{2\times \left(-7\right)}b^0}{a^{3}b^{-4}}=\frac{a^{-14}b^{0}}{a^{3}b^{-4}}=a^{-14-3}b^{0-(-4)}=a^{-17}b^{4}=\frac{b^{4}}{a^{17}}\)

Raise both sides of the equation to the inverse power of \(\displaystyle -3\) to cancel the exponent on the left hand side of the equation.

\(\displaystyle \rightarrow ((x+5)^{-3})^{-\frac{1}{3}} = (-1)^{-\frac{1}{3}}\)

\(\displaystyle \rightarrow x+5 = -1\)

Subtract \(\displaystyle 5\) from both sides:

\(\displaystyle \rightarrow (x+5) - 5 = (-1)-5\)

\(\displaystyle \rightarrow x = -6\)

\(\displaystyle \frac{x^{-12}}{x^{-4}y^{-8}}\)

Negative exponents are the reciprocal of their positive counterpart. For example:

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

Therefore:

\(\displaystyle \frac{x^{-12}}{x^{-4}y^{-8}} = \frac{x^4y^8}{x^{12}}\)

This simplifies to:

\(\displaystyle \frac{y^8}{x^8}\)

Rewrite the right-hand-side so that each side has the same base:

\(\displaystyle 2^{n+4}=2^{-6}\)

Use the Property of Equality for Exponential Functions:

\(\displaystyle n+4=-6\)

Solving for \(\displaystyle n\):

 \(\displaystyle n=-10\)

While a positive exponent says how many times to multiply by a number, a negative exponent says how many times to divide by the number.

To solve for negative exponents, just calculate the reciprocal.

\(\displaystyle 5^{-2}=\frac{1}{5^{2}}=\frac{1}{25}\)