Learning Assessment
Kushik Acharya
Presidential Business School
MTH 115: College Algebra
Ashwin Satyal
February 28, 2022
1.
Solution
a.
(-∞, ∞)
b.
(0, ∞)
2.
Solution
a.
Base = 1.08
y-intercept = (0,2)
Growth 8%
b.
Base = 0.87
y-intercept = (0,6)
Decay 13%
c.
Base = -
y intercept = (0,3)
Decay 7%
d.
Y = 23(1.12)^x
3.
Solution
Given points (-2,5) & (3,160)
We have, f(x) = ab^x
From 1st
point,
5 = ab^-2……i
From 2nd
point,
160 = ab^3……ii
Solving Eqn (i) and (ii), We get
a = 5/b^-2
using value of a in eqn ii we get,
160 = (5/b^-2)b^3
b = 2
using b in a = 5/b^-2
a= 20
Therefore, 20(2)^x is the required equation.
4.
Solution
a.
Consider 1984 as a base year. So we have,
(0, 1500) & (2, 4000)
f(x) = ab^x
f(0) = ab^o = 1500
a = 1500
f(2) = 1500b^2 = 4000
b = 1.63
so as asked by question, f(x) = 1500.(1.63)^t where t is years.
b.
In 1500 people which are getting infected from 1984 are increasing by a rate of 63%.
c.
If 1984 is considered as the base year then,
We have, 2001-1983= 17
So as asked by the question,
F(17) = 1500.(1.63)^17
= 6071283.795
5.
Solution
Given equation, 200(1.05)^t
As asked by question, we have
400 = 200(1.05)^t
2 = (1.05)^t
log(2) = t. log(1.05)
t= 14.2
Therefore, the doubling time is 14.2 years.
6.
Solution
a.
4(1.7)^x = 7(1.08)^x
(1.7)^x/(1.08)^x = 7/4
(1.574)^x = 1.75
log(1.574)^x = log (1.75)
x = 1.2336
b.
3e^x+5=7
lne(x+5) = 7/3
x = ln(7/3)-5
c.
log(x+3) = 3
x+3 = 10^3
x = 997
d.
log(x-1) + log(x+1) = 2
log((x-1)(x+1))= 2
log((x^2)-1)= 2
(x^2)-1= 10^2
x = √101
7.
Solution
As given by the question, the population doubles every 8 years.
a.
We have a formula for continuous growth rate which is given by,
f(t) = ae^k8
2a = ae^k8
2 = e^k8
ln(2) = lne^k8
8k= ln(2)
k = ln(2)/8
k= 0.8664
b.
Annual growth rate is given by,
e^k = 1+r
e^0.8664= 1+r
r= 0.090
Therefore, annual groth rate is 90%
8.
Solution
Alex = 4000(1+0.06/12)^12t
Shauna = 2000(1+0.09/4)^4t
As asked by the question we have to equate the values of Alex and Shauna
4000(1+0.06/12)^12t = 2000(1+0.09/4)^4t
t = -23.71
9.
Solution
a.
(½)a=a(1+r)^300
1+r= 300√1/2
r= -0.002307
Kryptonite is decaying at an annual rate of 2.307% per year.
b.
e^k= 1+r
k= -0.002309
Kryptonite is decaying continuously at 2.309%
10.
Solution
=2log(x)+3log(2)
=log(x)^2+log(2)^3
= log(x^2.2^3)
=log(8.x^2)
11.
Solution
y=2 as shown in the graph
considering two points (0,-1) & (1, -2)
f(x) = ab^x +2
-1 = ab^x +2
a = -1-2
a = -3
again,
-2 = -3b^1 + 2
-2-2 = -3b
b= 4/3
Therefore, required equation is f(x) = -3/(4/3)^x +2
12.
Solution
D- Linerar
A- Exponential
E- Exponential
C- Logarithmic
B- None of the above