Composed

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

reversed functions

We have seen that given an exponential function, we can reverse the pairs and get a logarithmic function. If the pairs of the exponential function look like $(D,R)$ , then the pairs of the logarithmic function look like $(R,D)$ . The same numbers are paired together. The just swap their roles between domain and range.

Composition

What if we were to compose these two functions? That is we make a new function whereby we take a domain number for the exponential function, get the function value, and then take this function value as the input into the logarithmic funciton, and get is output.

Setup

We have two partnered exponential and logarithmic functions: $E(x)$ and $L(t)$ .

(a): Let $d$ be a domain number of $E(x)$ .
(b): $d$ is in a pair: $(d, E(d))$ .
(c): View $E(d)$ as a number in the domain of $L(t)$ .
(d): We already know that $(E(d),d)$ is a pair in $L(t)$ , because $E$ and $L$ are partnered. Their pairs are reversed.
(e): That tells us that $L(t)=d$

$d$ was partnered with $E(d)$ . then we reversed the pair and $E(d)$ is partnered with $d$ .

The composition $L \circ E$ pairs $d$ with itself: $(d,d)$ for all of the domain numbers.

$(L \circ E)(x) = x$

Composition

We have seen that $E(x) = -2 \, \log _3(4-x)$ and $L(t) = 4 - 3^{-\frac {t}{2}}$ are two partnered exponential and logarithmic functions.

Map out the pairs for the composition, $(L \circ E)(d)$ .

Let $d$ be in the domain of $E(x)$ , then $E(d) = -2 \, \log _3(4-d)$ .

Now, let’s put this number into $L$ .

$(L \circ E)(d) = L(-2 \, \log _3(4-d)) = 4 - 3^{-\frac {-2 \, \log _3(4-d)}{2}}$

$(L \circ E)(d) = 4 - 3^{ \log _3(4-d)}$

$(L \circ E)(d) = 4 - \left ( \answer {4-d} \right )$

$(L \circ E)(d) = d$

Let’s try it the other way.

Composition

We have seen that $E(x) = -2 \, \log _3(4-x)$ and $L(t) = 4 - 3^{-\frac {t}{2}}$ are two partnered exponential and logarithmic functions.

Map out the pairs for the composition, $(E \circ L)(d)$ .

Let $d$ be in the domain of $L(t)$ , then $L(d) = 4 - 3^{-\frac {d}{2}}$ .

Now, let’s put this number into $E$ .

$(E \circ L)(d) = E(4 - 3^{-\frac {d}{2}}) = -2 \, \log _3(4 - (4 - 3^{-\frac {d}{2}}))$

$(E \circ L)(d) = -2 \, \log _3( 3^{-\frac {d}{2}})$

$(E \circ L)(d) = -2 \, \left ( \answer {-\frac {d}{2}} \right )$

$(E \circ L)(d) = d$

ooooo=-=-=-=-=-=-=-=-=-=-=-=-=ooOoo=-=-=-=-=-=-=-=-=-=-=-=-=ooooo
more examples can be found by following this link
More Examples of Percent Change