Logarithmic Functions

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exponents

An example of a basic exponential function is $E(t) = 2^t$.

Its graph looks like

We can evalute this function:

$E(0) = 2^0 = 1$
$E\left (\frac {1}{2}\right ) = 2^{\tfrac {1}{2}} = \sqrt {2}$
$E(-1) = 2^{-1} = \frac {1}{2}$

When we evaluate a function, we know the domain number and we seek its range partner. In this case, we know $d$ and we seek the pair $(d, 2^d)$, where $2^d$ is the value of the function at $d$.

Reverse

We can also think in reverse.

$E(t) = 4$
$E(t) = {\tfrac {1}{4}} $
$E(t) = 16 $

Here, we know the value of the function. We seek the domain numbers paired with it. We know $2^d$, we seek the pair $(d, 2^d)$, where $d$ is the solution to the equation.

$\bigstar $ For $E(t) = 2^t$, as long as we give a positive function value, then we can find the associated domain number.

$\bigstar $ For $E(t) = 2^t$, every positive function value has exactly one associated domain number. That is eeriely backwards of the function rule.

Backwards

Remember: Each domain number in a function is paired with exactly one range number. In other words, a function can have only one value at each domain number.

We just said a similar sentence for the range numbers of an exponential function.

We just said that each range number for an exponential function is paired with exactly one domain number.

That’s not a requirement to be a function. It is an extra characteristics of exponential functions. This range characteristic is called one to one.

Thinking of an exponential function in reverse sounds like a new function. It is. We call it a logarithmic function.

\[ \begin{array}{lcl} Exponential Function & & Logarithmic Function \\ domain = (-\infty , \infty ) & \searrow & domain = (0, \infty ) \\ range = (0, \infty ) & \nearrow & range = (-\infty , \infty ) \\ (a, r^a) & \longrightarrow & (r^a, a) \end{array} \]

The logarithm function just reverses the pairs in the exponential function. If $(a, 2^a)$ is a pair in the exponential function, then $(2^a, a)$ is a pair in the logarithmic function.

Therefore, the domain and range switch.

Logarithm

It seems weird to write $(2^a, a)$, even though it is perfectly correct.

We are used to representing the domain number by a letter, $b$, and then the function value as a formula involving $b$.

We would prefer our pairs in the logarithm function to look like

\[ (b, expression) \]

What’s the expression?

Logarithmic functions come from the study of logarithms, which gets shortened to log for notation purposes.

In our example here, we are working with the base $2$ exponential function, $E(t) = 2^t$. So, the reverse is called the logarithm base $2$. We tack on a subscript $2$ to complete the name of our new function.

\[ \log _2(b) \]

That pairs in the logarithmic base $2$ function look like

\[ (b, \log _2(b)) \]

The coordinates of the points on the graph look like

\[ (b, \log _2(b)) \]

$\log _2(b)$ is just the exponent that $2$ needs to equal $b$.

$\blacktriangleright $ Remember: $(b, \log _2(b))$ are the reverse of the exponential pairs : $(a, 2^a)$.

$(b, \log _2(b))$ is also $(2^c, c)$ for some $c$.

$\blacktriangleright $ $\log _2(b)$ is the number that you raise $2$ to, to get $b$.

\[ 2^{\log _2(b)} = b \]

We have a logarithmic function for every exponential function. They are designated by their bases.

Logarithm Function base $r$ : $log_r(t)$

$\log _r(t)$ is the number you raise $r$ to, to get $t$.

$t \in (0, \infty )$.

$\log _r(t) \in (-\infty , \infty )$.

Since all of the pairs are reversed, the graphs switch axes.

The horizontal axis is an asymptote for the basic exponential function. Now, the vertical axis is an asymptote for the basic logarithmic function.

The graph of a basic exponential function has $(0,1)$ as an intercept. The graph of a basic logarithmic function has $(1,0)$ as an intercept.

Their graphs are mirror images of each other across the diagonal through quadrants I and III.

Logarithmic functions don’t really have vertically shifted versions. We can talk about a basic logarithm function being shifted.

\[ L(x) = A \log _r(B x + C) + D \]

But, the $D$ can be explained with additional horizontal transformations.

We know that the range of $\log _r(x)$ is all real numbers, which includes $D$. Therefore, there must be a real number, $d$, such that $\log _r(d) = D$.

That gives us

\[ L(x) = A \log _r(B x + C) + D = A \log _r(B x + C) + \log _b(d) \]

And, we have logarithm rules, which give us

\[ L(x) = A \log _r(B x + C) + D = A \log _r((B x + C)d) = A \log _r(B d x + C d) \]

Any shifted logarithmic function can be rewritten as a logarithmic function. SO, we might as well call all of them logarithmic functions.

Therefore, we separate exponential and shifted exponential, but we don’t separate logarithmic and shifted logarithmic.

Note: Many times people just speak about exponential functions, and they mean both exponential and shifted exponential.

From this basic logarithmic function and its graph we can analyze and transform for more general logarithmic functions.

Logarithmic Function

Analyze $M(t) = \log _2(t+3) - 4$

$\blacktriangleright $ The inside of the logarithm is $\answer {t+3}$ and this equals $0$ when $t=-3$. This has to be the vertical asymptote.

$\blacktriangleright $ The inside of the logarithm is $t+3$, and this is positive for $t>-3$. The graph must move to the right.

$\blacktriangleright $ The leading coefficient is $\answer {1}$, which is positive. Therefore, the graph hugs the asymptote up down the asymptote.

$\blacktriangleright $ $t+3=1$ when $t=-2$. Therefore, the anchor point $(1,0)$ has moved to $t = -2$. $M(-2) = -4$. This gives us the point $\left ( \answer {-2}, \answer {-4} \right )$.

Graph of $y = M(t)$.

Our analysis tells us that:

The implied domain of $M$ is $(-3,\infty )$.
The implied range of $M$ is $\mathbb {R}$.
$M$ is always increasing decreasing.
$M$ has no maximums or minimums.
$\lim \limits _{t \to -3^+} M(t) = -\infty $
$\lim \limits _{t \to \infty } M(t) = \infty $

We can also see that the graph will have a horizontal intercept, which means the function has a zero.

$M(t) = \log _2(t+3) - 4 = 0$

\begin{align*} \log _2(t+3) - 4 & = 0 \\ \log _2(t+3) & = 4 \\ 2^{\log _2(t+3)} & = 2^{\answer {4}} \\ t+3 & = 16 \\ t & = 13 \end{align*}

$\blacktriangleright $ Remember: $\log _2(t+3)$ is the thing that you raise $2$ to, to get $t+3$ and $\log _2(t+3) = 4$.

Therefore, $4$ is the thing that you raise $2$ to, to get $t+3$. $t+3$ must be $16$.

$M(t)$ is a logarithmic function, so it must have a partner (shifted) exponential function. The pairs for $M$ look like $(t, M(t))$ or just $(t,M)$. The pairs for the (shifted) exponential function would look like $(M, t)$. The roles of $M$ and $t$ would be switched. $M$ would be the variable in the formula. $t$ would be the function value.

We can obtain the formula for this partner (shifted) exponential function by solving the logarithmic formula for $t$.

\begin{align*} \log _2(t+3) - 4 & = M \\ \log _2(t+3) & = M + 4 \\ 2^{\log _2(t+3)} & = 2^{\answer {M+4}} \\ t+3 & = 2^{M+4} \\ t & = \answer {2^{M+4} - 3} \end{align*}

$\blacktriangleright $ Remember: $\log _2(t+3)$ is the thing that you raise $2$ to, to get $t+3$ and $\log _2(t+3) = M+4$.

Therefore, $M+4$ is the thing that you raise $2$ to, to get $t+3$

Here is the graph of both the logarithmic and the associated exponential functions.

Logarithmic Function

Analyze $K(x) = -2 \, \log _3(4-x)$

$\blacktriangleright $ The inside of the logarithm is $4-x$ and this equals $0$ when $x=\answer {4}$. $x=4$ has to be the vertical asymptote.

$\blacktriangleright $ The inside of the logarithm is $4-x$, and this is positive for $x<4$. The graph must move to the right.

$\blacktriangleright $ The leading coefficient is $-2$, which is positive negative . Therefore, the graph is flipped vertically from the basic graph. It goes up the asymptote.

$\blacktriangleright $ $4-x=1$ when $x=3$. Therefore, the anchor point $(1,0)$ has moved to $\left ( \answer {3}, \answer {0} \right )$.

Graph of $y = K(x)$.

The natural or implied domain of $K$ is $(-\infty , 4)$.
The implied range of $K$ is $\mathbb {R}$.
$K$ is always increasing.
$K$ has no maximums or minimums.
$\lim \limits _{x \to 4^-} K(x) = \infty $
$\lim \limits _{x \to -\infty } K(x) = -\infty $

Reversing all of the pairs will give the associated exponential function.

$K(x)$ is a logarithmic function, so it must have a partner exponential function. The pairs for $K$ look like $(x, K)$. The pairs for the exponential function would look like $(K, x)$. The roles of $K$ and $x$ would be switched. $K$ would be the variable in the exponential formula. $x$ would be the exponential function value.

We can obtain the formula for this partner exponential function by solving the logarithmic formula for $x$.

\begin{align*} -2 \,\log _3(4-x) & = M \\ \log _3(4-x) & = \answer {-\frac {M}{2}} \\ 3^{\log _3(4-x)} & = \answer {3}^{-\frac {M}{2}} \\ 4 - x & = 3^{-\frac {M}{2}} \\ \answer {4 - 3^{-\frac {M}{2}}} & = x \\ \end{align*}

$\blacktriangleright $ Remember: $\log _3(4-x)$ is the thing that you raise $3$ to, to get $4-x$ and $\log _3(4-x) = - \frac {M}{2}$.

Therefore, $- \frac {M}{2}$ is the thing that you raise $3$ to, to get $4-x$

Here is the graphs of both the logarithmic and the associated exponential functions.

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2025-01-07 02:20:01