how to graph logarithmic functions

a logarithmic function in the form $f(x)=a\cdot {\log }_c(b(x-h))+k$ can be graphed

they have a way more restricted domain compared to exponential functions because the logarithm function has restraints - for ${\log }_a(b)$, the base $a$ is restricted to be $0<a<1\vee a>1$ and the input, $b$, is restricted to be $b>0$

because of the restraints on the log function, we know that $b(x-h)>0$ is the restraint on the domain. as soon as $b(c-h)=0$, the function breaks. therefore, the asymptote line is $x=m$ where $b(m-h)=0$

assume that $a$ and $b$ are positive and $c>1$, remember how a general logarithmic function looks like
yes, since a log is the inverse of an exponent, it’s a flip across the line $y=x$

figure: the general logarithmic function goes from the bottom to the upper right

now, we think about other values for $a$, $b$ and $c$

if a is negative, there will be a vertical reflection
if $b$ is negative, there will be a horizontal reflection across the asymptote
if $c$ is in the range $0<c<1$, there will be a vertical reflection

worked example: graph $y=-{\log }_4(2x-1)$
simplify to the form $f(x)=a\cdot {\log }_c(b(x-h))+k$
$y=-1\cdot {\log }_4(2(x-\frac{1}{2}))$

asymptote is $x=m$ where $b(m-h)=0$
$2(m-\frac{1}{2})=0$
$m=\frac{1}{2}$
asymptote is $x=\frac{1}{2}$

x-intercepts

$0=-{\log }_4(2x-1)$
$0={\log }_4(2x-1)$
$4^0=2x-1$
$x=1$

y-intercepts

$y=-{\log }_4(-1)$

the input to the log function is not positive
y-intercept is undefined

worked example: graph $y=\frac{1}{2}{\log }_4(x+1)-1$
simplify to the form $f(x)=a\cdot {\log }_c(b(x-h))+k$
$y=\frac{1}{2}\cdot {\log }_4(x+1)-1$