Domain: $⟨-4,\to ⟩$
Range: $ℝ$

$x=-4$

First shift $-4$ in the $x$ -direction, then multiply by $-3$ along the $y$ -axis, and finally shift $1$ in the $y$ -direction.

$1-3\cdot log(x+4)=0$ gives you $log(x+4)=\frac{1}{3}$ and $x+4={10}^{\frac{1}{3}}=\sqrt[3]{10}$, so $x=\sqrt[3]{10}-4$.
The intersect is $(\sqrt[3]{10}-4,0)$.

${}^{\frac{1}{2}}log\left(x\right)=3$, so $x={\left(\frac{1}{2}\right)}^3=\frac{1}{8}$.

${}^{2}log\left(x\right)=-3$, so $x=2^{-3}=\frac{1}{8}$.

$(\frac{1}{8},-3)$

For example $(2,-1)$ and $(2,1)$.

$h\left(x\right)=k\left(x\right)$ if $x=1$.

${}^{\frac{1}{2}}log\left(x\right)=\frac{log\left(x\right)}{log\left(\frac{1}{2}\right)}=\frac{log\left(x\right)}{log\left(2^{-1}\right)}=\frac{log\left(x\right)}{\mathrm{-}log\left(2\right)}=\mathrm{-}\frac{log\left(x\right)}{log\left(2\right)}$
${}^{2}log\left(x\right)=\frac{log\left(x\right)}{log\left(2\right)}$
This shows that ${}^{\frac{1}{2}}log\left(x\right)=\mathrm{-}{}^{2}log\left(x\right)$.

$21=1+a\cdot log\left(100\right)$ gives you $21=1+2a$ and therefore $a=10$.

See graph.

$31=1+10\cdot log\left(x\right)$ gives you $log\left(x\right)=3$ and therefore $x=1000$. The answer is thus $1000$ ASA.

${\text{D}}_f=⟨0,\to ⟩$, ${\text{B}}_{\left(f\right)}=ℝ$, vertical asymptote $x=0$.
${\text{D}}_{\left(g\right)}=⟨\leftarrow ,2⟩$, ${\text{B}}_g=ℝ$, vertical asymptote $x=2$.

First mirror the graph on the $y$ -axis (or multiply by $-1$ along the $x$ -axis) and then shift $2$ units in the $x$ -direction.

$x=2-x$ gives you $x=1$.

The vertical line $x=1$.