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Update Unit 1: Exponential and Logarithmic Functions.md
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A function is a relation where each x-value maps to exactly one y-value.
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If given a function in the form $`y = af[k(x-d)] + c`$, then let $`(x,y)`$ be the original points, the new points will be $`(\dfrac{1}{k}x+d, ay+c)`$.
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If given a function in the form $`y = af[k(x-d)] + c`$, then let $`(x,y)`$ be the original points, the new points will be $`(\dfrac{1}{k}x+d, ay+c)`$.
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The domain and range of the exponential function is:
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- $`D : \{x | x \in \mathbb{R}\}`$
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- $`R : \{y | y > 0, y \in \mathbb{R}\}`$
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The domain and range of the logarithmic function is
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- $`D: \{x | x > 0, x \in \mathbb{R}\}`$
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- $`R: \{y | y \in \mathbb{R}\}`$
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If $`f(x)`$ is a function, then the inverse is $`f^{-1}(x)`$. The inverse has the following properties:
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- Domain of $`f(x)`$ = Range of $`f^{-1}(x)`$
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- Range of $`f(x)`$ = Domain of $`f^{-1}(x)`$
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Graphically, the inverse of a function is by reflecting the original function over the line $`y=x`$.
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A vertical line test is used to test whether a relation is a function. If any 2 points can be drawn through a vertical line, then that relation is **not a function**.
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To solve/find the inverse of a function, just swap the $`y`$ and $`x`$ and isolate/solve for $`y`$.
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## Exponential Decay/Growth
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When the base ($`b`$) is in the range $`0 \lt b \lt 1`$, the exponential funciton is said to have a **exponential decay**, the smaller the base, the stronger the decay.
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When the base ($`b`$) is in the range $`b \gt 1`$, the exponential function is said to have a **exponential growth**, the bigger the base, the stronger the growth.
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## Graphing Exponential Functions
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If you have **exponential growth** (meaning your base is greater than $`1`$), use more positive values rather than negative values.
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If you have **exponential decay** (meaning your base is in the range $`(0, 1)`$), use more negative values rather than positive values.
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**Don't forget the asymtote.**
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## Logarithmic Function
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The logarithmic function is the **inverse** of the exponential function.
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In essence, if $`x = b^y`$, then $`\log_b x = y`$
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**Note:** The logarithm is defined only for $`b > 0, b \ne 1`$
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**Note 2:** The symbol $`ln`$ is $`log_e`$, we usually call it the **natural log**.
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## Logrithm Laws
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1. $`\log_b(b^x) = x`$
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2. $`b^{\log_b(x)} = x`$
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3. $`\log_b(1) = 0`$
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4. $`\log_b(b) = 1`$
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|Law|Form|Example|
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|:--|:---|:------|
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|Change Of Base (COB)|$`\log_a(b) = \dfrac{\log_m(b)}{\log_m(a)}`$|$`\log_2(5) = \dfrac{\log_{10}(5)}{\log_{10}(2)}`$|
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|Change Of Base (COB)|$`\log_a(b) = \dfrac{1}{\log_b(a)}`$|$`\log_2(5) = \dfrac{1}{\log_5(2)}`$|
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|Power Law|$`\log_b^m(x^n) = \dfrac{n}{m}\log_b(x)`$|$`\log_{2^2}(4^3) = \dfrac{3}{2}\log_2(4)`$|
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|Product Law|$`\log_b(xy) = \log_b(x) + log_b(y)`$|$`\log_2(2 \times 3) = \log_2(2) + \log_2(3)`$|
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|Quotient Law|$`\log_b(\dfrac{x}{y}) = \log_b(x) - \log_b(y)`$|$`log_2(\dfrac{2}{3}) = \log_2(2) - \log_2(3)`$
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## Solving Logarithms
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1. Using a common base and equating the the 2 exponents to one another ($`2^x = 4^{x-5} \implies 2^x = 2^{2x-10} \implies x = 2x-10`$)
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2. Using a log rule to simplify and bring the exponenets to the "living room/main floor".
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## Application of Exponential Growth
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The formula for **Exponential Growth** is given as:
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```math
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\LARGE
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N = N_0(R)^{\frac{t}{d}}
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```
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$`N = `$ Final amount.
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$`N_0 = `$ Starting amount.
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$`R =`$ Growth factor.
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- $`R = 1 + r`$
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- **half-life:** $`R = \dfrac{1}{2}`$
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- **doubling time:** $`R = 2`$
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**Growth Rate**
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- $`r > 0`$ Exponential Growth
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- $`-1 \lt r \lt 0`$ Exponential Decay
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- r is ually given as a $`\%`$
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$`t = `$ Total amount. (time for $`N_0`$ to get to $`N`$)
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$`d = `$ Growth Rate time. (Time for 1 Growth Rate to occur).
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