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highschool/Grade 10/Math/MCR3U7/Unit 1: Exponential and Logarithmic Functions.md

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Unit 1: Exponential and Logarithmic Functions

Review

Function: A relation where each x-value maps to exactly one y-value.

If given a function in the form y=af[k(xd)]+c`y = af[k(x-d)] + c`, then let (x,y)`(x,y)` be the original points, the new points will be (1kx+d,ay+c)`(\dfrac{1}{k}x+d, ay+c)`.

The domain and range of the exponential function is: - D:{xxR}`D : \{x | x \in \mathbb{R}\}` - R:{yy>0,yR}`R : \{y | y > 0, y \in \mathbb{R}\}`

The domain and range of the logarithmic function is - D:{xx>0,xR}`D: \{x | x > 0, x \in \mathbb{R}\}` - R:{yyR}`R: \{y | y \in \mathbb{R}\}`

If f(x)`f(x)` is a function, then the inverse is f1(x)`f^{-1}(x)`. The inverse has the following properties: - Domain of f(x)`f(x)` = Range of f1(x)`f^{-1}(x)` - Range of f(x)`f(x)` = Domain of f1(x)`f^{-1}(x)`

Graphically, the inverse of a function is by reflecting the original function over the line y=x`y=x`.

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.

To solve/find the inverse of a function, just swap the y`y` and x`x` and isolate/solve for y`y`.

Exponential Decay/Growth

When the base (b`b`) is in the range 0<b<1`0 \lt b \lt 1`, the exponential function is said to have a exponential decay, the smaller the base, the stronger the decay.

When the base (b`b`) is in the range b>1`b \gt 1`, the exponential function is said to have a exponential growth, the bigger the base, the stronger the growth.

Graphing Exponential Functions

If you have exponential growth (meaning your base is greater than 1`1`), use more positive values rather than negative values.

If you have exponential decay (meaning your base is in the range (0,1)`(0, 1)`), use more negative values rather than positive values.

Dont forget the asymtote.

Logarithmic Function

The logarithmic function is the inverse of the exponential function.

In essence, if x=by`x = b^y`, then logbx=y`\log_b x = y`

Note: The logarithm is defined only for b>0,b1`b > 0, b \ne 1`

Note 2: The symbol ln`ln` is loge`log_e`, we usually call it the natural log.

Logrithm Laws

  1. logb(bx)=x`\log_b(b^x) = x`
  2. blogb(x)=x`b^{\log_b(x)} = x`
  3. logb(1)=0`\log_b(1) = 0`
  4. logb(b)=1`\log_b(b) = 1`
Law Form Example
Change Of Base (COB) loga(b)=logm(b)logm(a)`\log_a(b) = \dfrac{\log_m(b)}{\log_m(a)}` log2(5)=log10(5)log10(2)`\log_2(5) = \dfrac{\log_{10}(5)}{\log_{10}(2)}`
Change Of Base (COB) loga(b)=1logb(a)`\log_a(b) = \dfrac{1}{\log_b(a)}` log2(5)=1log5(2)`\log_2(5) = \dfrac{1}{\log_5(2)}`
Power Law logbm(xn)=nmlogb(x)`\log_b^m(x^n) = \dfrac{n}{m}\log_b(x)` log22(43)=32log2(4)`\log_{2^2}(4^3) = \dfrac{3}{2}\log_2(4)`
Product Law logb(xy)=logb(x)+logb(y)`\log_b(xy) = \log_b(x) + log_b(y)` log2(2×3)=log2(2)+log2(3)`\log_2(2 \times 3) = \log_2(2) + \log_2(3)`
Quotient Law logb(xy)=logb(x)logb(y)`\log_b(\dfrac{x}{y}) = \log_b(x) - \log_b(y)` log2(23)=log2(2)log2(3)`log_2(\dfrac{2}{3}) = \log_2(2) - \log_2(3)`

Solving Logarithms

  1. Using a common base and equating the the 2 exponents to one another (2x=4x5    2x=22x10    x=2x10`2^x = 4^{x-5} \implies 2^x = 2^{2x-10} \implies x = 2x-10`)
  2. Using a log rule to simplify and bring the exponenets to the “living room/main floor”.

Application of Exponential Growth

The formula for Exponential Growth is given as:

\LARGE

N = N_0(R)^{\frac{t}{d}}

N=`N = ` Final amount

N0=`N_0 = ` Starting amount

R=`R =` Growth factor - R=1+r`R = 1 + r` - half-life: R=12`R = \dfrac{1}{2}` - doubling time: R=2`R = 2`

Growth Rate - r>0`r > 0`: Exponential growth - 1<r<0`-1 \lt r \lt 0`: Exponential decay - r is usually given as a percentage (%`\%`)

t=`t = `: Total time for N0`N_0` to get to N`N`

d=`d = ` Time for 1 growth rate to occur