MAT4MDS Assignment 2: Equations, Graphs, and Function Analysis

This document provides a comprehensive solution to MAT4MDS Assignment 2, encompassing various mathematical concepts. The assignment begins with solving several equations involving logarithmic and exponential functions, followed by mathematical proofs demonstrating fundamental logarit...Read More

Order Id 953574 MAT4MDS ASSIGNMENT 2, 2019
Question 1
a) Solving these equations for x
i) loge(loge(3x)) = 0 applying the log rule 0 = ln(e0) = ln(1)
loge(loge(3x)) = ln(1)
loge(3x) = 1
e1 = 3x ⟹ 𝐱 =
𝐞𝟏
𝟑
ii) ex2−5x+6 = 1
ln(ex2−5x+6) = ln(1)
x2 − 5x + 6 = 0solving the quadratic equation
(x − 3)(x − 2) = 0
∴ 𝐱 = 𝟑 , 𝐱 = 𝟐
iii) log3(x2 − 9) − log3(x + 3) = 2 − log3(x + 1)
log3(x2 − 9) − log3(x + 3) + log3(x + 1) = 2
log3 ( (x2 − 9)(x + 1)
(x + 3) ) = 2
32 = (x + 3)(x − 3)(x + 1)
(x + 3) ⟹(x − 3)(x + 1) = 9
x2 − 2x − 12 = 0
∴ 𝐱 = 𝟏 +√𝟏𝟑 𝐎𝐫 𝐱 = 𝟏 +√𝟏𝟑
𝐱 = 𝟒. 𝟔𝟎𝟓𝟓𝟓𝟏𝟐𝟕𝟓 𝐨𝐫 𝐱 = −𝟐. 𝟔𝟎𝟓𝟓𝟓𝟏𝟐𝟕𝟓in decimal format
iv) 9𝑥 − 2.3𝑥+1 = 7

Using trial and error method
𝒙 = 𝟏. 𝟏𝟕𝟎𝟖𝟑
b) Proof
i) loga(b) logb(a) = 1
logb(a) = 1
loga b
1 = 1
loga b × 1
logb a × (loga(b) logb(a))
1 = ( 1
loga b × loga(b)) × ( 1
logb a × logb(a))
1 = 1 × 1 = 1
ii) log1
b
(x) = − logb(x)
logb−1(x) = − logb(x) use of change of base rule
ln(x)
ln(b−1) = − ln(x)
ln(b)
ln(x)
−ln(b) = − ln(x)
ln(b)
ln(x)
ln(b) = − ln(x)
ln(b)
c) Graphing the following equations on the same axis, c=8
𝑦 = 𝑥, (1
5)
𝑥
, (1
8)
𝑥

d) Estimating the slope of the red line
The graph is logarithmic on y-axis
s = ∆y
∆x= ∆W
∆n at(1,1) and (2,10)
Where W total number of websites, and n the number of year
s= ∆W
∆n = 1
2−1 = 1 ∆W = 1for one log cycle
∴ s = 1
yintercept = −10
Thus, y = −10(10)x
The relationship between W and n is Exponential relationship
Question 2
F:[0, 1} → ℝ F(x) = 1 −(1 − xa)b

a) The range for F is [0, 1]
b) F(x) = 1 −(1 − xa)b
f = 1 − xa g = xb
c) a = 5, b ==
1
8
𝑔°𝑓(𝑥) = (1 − 𝑥𝑎)1
8
Reflecting on x-axis
Shifting vertically up by +1

d) Both a and b are scale parameters since they both change the size and the shape of the
function.
e) F−1(x)
y = 1 −(1 − xa)b
(1 − xa)b = 1 − y
ln(1 − xa) = ln(1 − y)
b
1 − xa = eln(1−y)
b ⟹ xa = 1 − e
ln(1−y)
b ⟹ aln(x) = ln (1 − e
ln(1−y)
b )
x = e
ln(1−e
ln(1−y)
b )
a
∴ F−1(x) = e
ln(1−e
ln(1−x)
b )
a
Range:[0, 1]
Domain: [0, 1]