The inverse function of $y=a^x$ is $y=\log_{a}{x}$. Therefore $y=\log_{a}{x}$ is an inverse function, it is a reflection of $y=a^x$ in the line $y=x$.
\begin{array}{|c|c|c|} \require{AMSsymbols} \require{color} \hline
& y=a^x & \color{red}y =\log_{a}{x} \\ \hline
\text{domain} & x \in \mathbb{R} & \color{red}x \gt 0 \\ \hline
\text{range} & y \gt 0 & \color{red}y \in \mathbb{R} \\ \hline
\text{asymptote} & horizontal\ y=0 & \color{red}vertical\ x=0 \\ \hline
\text{fixed point} & (0,1) & \color{red}(1,0) \\ \hline
\end{array}
Example 1
(a) Sketch the graphs of $y=\log_{2}{x}$ and $y=\log_{2}{(x+1)}$.
Shift to the left side by $1$ unit.
(b) Sketch the graphs of $y=\log_{2}{x}$ and $y=\log_{2}{x-2}$.
Shift down by $2$ units.
(c) Sketch the graphs of $y=\log_{2}{x}$ and $y=\log_{2}{(x+1)-2}$.
Shift to the left side by $1$ and down by $2$ units.
(d) Find the domain of $y = \log_{2}{(x+1)}-2$.
\( \begin{align} \displaystyle
x + 1 &\gt 0 \\
\therefore x &\gt -1 \\
\end{align} \)
(e) Find the range of $y = \log_{2}{(x+1)}-2$.
$x \in \mathbb{R}$ or all real $x$
(f) Find any asymptote(s) of $y = \log_{2}{(x+1)}-2$.
\( \begin{align} \displaystyle
x + 1 &= 0 \\
\therefore x &= -1 \\
\end{align} \)
(g) Find any $x$-intercept(s) of $y = \log_{2}{(x+1)}-2$.
\( \begin{align} \displaystyle \require{color}
\log_{2}{(x+1)}-2 &= 0 &\color{red}y=0 \\
\log_{2}{(x+1)} &= 2 \\
x+1 &= 2^2 \\
x+1 &= 4 \\
x &= 3 \\
\therefore (3,0) \\
\end{align} \)
(h) Find any $y$-intercept(s) of $y = \log_{2}{(x+1)}-2$.
\( \begin{align} \displaystyle
y &= \log_{2}{(0+1)}-2 &\color{red}x=0 \\
y &= \log_{2}{1}-2 \\
y &= 0-2 \\
y &= -2 \\
\therefore (0,-2) \\
\end{align} \)
Example 2
Sketch the graphs of $y=\log_{2}{x}$ and $y=3\log_{2}{x}$.
Example 3
Sketch the graphs of $y=\log_{2}{x}$ and $y=\log_{2}{(3x)}$.
Example 4
Sketch the graphs of $y=\log_{2}{x}$ and $y=-\log_{2}{x}$.
Example 5
Sketch the graphs of $y=\log_{2}{x}$ and $y=\log_{2}{(-x)}$.
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