Maths Assignment

This document is a Maths Assignment that includes various questions and their solutions. It covers topics such as logarithms, exponential growth, and population growth. The assignment provides step-by-step solutions to each question.

Running head: MATHS ASSIGNMENT 1
Math Assignment
Professor’s Name:
Name:
Date:

MATHS ASSIGNMENT 2
Question 1
a) log A=log B+ log C
log A=log (BC)
A=BC
b) loga F=loga 3−loga c
loga F+loga c=loga 3
loga ( Fc)=loga 3
Fc=3
c) log C=−1
3 log g
log C=log(g
−1
3 )
C=g
−1
3
d) log Q=2 log p+1
log Q=log p2+ log10
¿ log 10 p2
Q=10 p2
Question 2
a) 50 ×e−0.03 x=0.05
e−0.03 x=0.05
50 =0.001
ln e−0.03 x=ln 0.001
−0.03 x=ln0.001
x= ln 0.001
−0.03
b) 81 e0.3 x−27=0
81 e0.3 x=27
e0.3 x=27
81
e0.3 x=1
3
ln e0.3 x=ln 1
3
0.3 x=ln 1
3
x=
ln 1
3
0.3

MATHS ASSIGNMENT 3

MATHS ASSIGNMENT 4
Question 3
a) log(10× 3
√ 10 ¿)¿
¿ log( 10× 10
1
3 )
¿ log 10
4
3
¿ 4
3 log 10
¿ 4
3 ( 1 )
¿ 4
3
b) log ( 100
√10 )
¿ log (102 ×10
−1
2 )
¿ log ( 10
3
2 )
¿ 3
2 log 10
¿ 3
2 ( 1 )
¿ 3
2
c) ln ( ea
eb ) ¿ ln ( ea−b )
¿ ( a−b ) ln e
¿ ( a−b ) ( 1 )
¿ a−b
d) loga ( 1
√a )
¿ loga ( a
−1
2 )
¿−1
2 loga a
−1
2 (1)
¿−1
2
e) log10
4
√ 1000

MATHS ASSIGNMENT 5
¿ log10 1000
1
4
¿ log10 10
3
4
¿ 3
4 log10 10
¿ 3
4
f) log 4
3 + log3+ log7
¿ log ( 4
3 × 3 ×7 )
¿ log 28
g) 2− 1
2 logn 4−logn 5
¿ 2−logn 4
−1
2 −logn 5
¿ 2−logn 2−logn 5
¿ 2−logn 10
Question 4
The growth rate for an exponential growth is given by the formula:
k = ln 2
doubling time , where k = growth rate
¿ ln 2
44
¿ 0.01575
The growth rate = 0.01575 per hour
Question 5
Doubling time = ln 2
k
¿ ln 2
0.03/hour =23.10 hours
The doubling time = 23.10 hours
Question 6
The population at a given time (Pt) is given by the formula:

MATHS ASSIGNMENT 6
Pt =P0 ekt
Where P0 is the initial population at time t=0, k = growth rate and t= elapsed time
In this case: P0 = 400 million, k = 0.08 per hour, t = 7 hours
Pt =400 × e(0.08 ×7)
¿ 700.27
The population after 7 hours is 700.27 million

MATHS ASSIGNMENT 7
Question 7
Amount remaining after t years (At) is given by the formula:
At = A0 e−kt
Where A0 is the initial population at time t=0, k = decay rate and t= elapsed time
In this case: 0.02 decays every 1000 years. Therefore, k = 0.02
1000 =0.00002 per year
So we have: k = 0.00002 per year, A0 = 700 units, t = 2000 years
At =700 e−(0.00002 × 2000)
¿ 672.55
The units remaining after 2000 years will be 672.55 units.