MATH 115 Quiz 3 Summer 2018

This article provides solutions to MATH 115 Quiz 3 Summer 2018, covering topics such as exponential equations, logarithms, and compound interest problems. The solutions include step-by-step explanations and calculations. The course code, course name, and college/university are not mentioned.

MATH 115 Quiz 3 Summer 2018
1. Solve the following: 21−x=42 x
Solution: 21−x=42 x
⟹ 21−x= ( 22 )
2 x
⟹ 21−x=24 x
If base term is same then power will be equal.
Therefore, ⟹ 1−x=4 x
⟹ 1=5 x
⟹ x= 1
5
2. How long will it take an investment to double if it is invested at 7.8%, compounded
continuously?
Solution: Given r = 7.8 % = 7.8
100 = 0.078
if the interest rate is compounded continuously, then the general form is:
a=b ert
a → Amount after time ‘t’
b → Initial amount
r → Rate of interest compounded continuously
t → Time in years
Let, Initial amount = b
Amount after time ‘t’ = 2b (doubled)
Then equation looks like
⟹ 2 b=b e0.078 t
⟹ 2=e0.078 t
⟹ 0.078 t=ln 2

⟹ 0.078 t =0.693147
⟹ t= 0.693147
0.078
⟹ t=8.8865 Years
3. An initial investment of $1000 is appreciated for 2 years in an
account that earns 6% interest, compounded semiannually.
Find the amount of money in the account at the end of the
period.
Solution: Given
Initial investment P = $ 1000
Time ‘ t ‘ = 2 years
Rate of interest = r = 6 % = 0.06
Semiannually = N = 2
Let us suppose at the end of period, amount of money be A
Then A=P(1+ r
N )
t N
Where A=1000 (1+ 0.06
2 )2∗2
A=1000 ( 1+ 0.03 ) 4
A=1000 (1.03 )4
A=$ 1125.5
4. A company begins a radio advertising campaign in Chicago to
market a new soft drink. The percentage of the target market
that buys a soft drink is estimated by the function
Solution:
Given:
P ( t ) =100−100 e−0.01 t
Where P(t) is the percentage of the target market that buys soft drink.
“t” is the number of days of campaign.

Given, P ( t ) =75 %
⟹ 75=100−100 e−0.01t
⟹−25=−100 e−0.01 t
⟹ 1=4 e−0.01 t
⟹ 1
4 =e−0.01t
Take natural log both sides, ⟹ ln 1
4 =−0.01t
⟹ ln 0.25=−0.01t
⟹−1.38629=−0.01 t
t=138.63 days
5. A certain radioactive isotope decays at a rate of 0.15%
annually. Determine the half-life of this isotope, to the nearest
year.
Solution:
Radioactive isotope decay takes place exponentially
N=N 0 ekt (1)
Where N is the concentration of isotope left after time ‘t’
and No is the Initial concentration
k is the order of decay
Therefore,
Given , t = 1 year
Concentration of isotope left after 1year,
N=N 0−( 0.15 % of N0 )
N=N 0− 0.15
100 N0
N=N 0 (1−0.0015)

N=0.9985 N 0
Now plugging “N” and “t” value in the exponential equation (1)
N=N 0 ekt
⟹ 0.9985 N 0=N0 ek∗1
⟹ 0.9985=ek
Now taking natural log both sides, we have
⟹ ln 0.9985=ln ek
⟹ ln 0.9985=k∗ln e
⟹ ln 0.9985=k ∴ ln e=1
k =−0.0015
Now conditions at half-life period (t1/2)
at t1
2
N = N 0
2
Therefore,
N=N 0 ekt becomes
⟹ N0
2 =N 0 e
k t 1
2
⟹ 1
2 =ek t 1
2
Taking natural log both sides, we have
⟹ ln 0.5=ln ekt 1
2
⟹ ln 0.5=k t1
2
∗ln e ∴ ln e=1
Then,

⟹−0.69314=−0.0015 t1
2
⟹ t1
2
=462.098 years
6. The population of a small country increases according to the
function B ( t )=1900000e0.04 t where t is measured in years. How many
people will the country have after 8 years?
Solution: The population of small country increases according to the function equation:
B ( t )=1900000e0.04 t ∴ where t is∈ years
After 8 years, t=8
B ( 8 )=1900000 e0.04∗8
B ( 8 )=1900000 ×1.377
B ( 8 )=2616300
Population after 8 years will be 2616300.
7. Find the value of the expression.
ln ( e−2 )
Solution: Let y=ln ( e−2 )
y=−2∗lne
y=−2∴ ln e=1
8. Express as a single logarithm and, if possible, simplify
1
2 loga x+ 4 loga y−3 loga x
Solution:

We see that base of the logarithm is same i.e “a”
¿ 1
2 loga x+ 4 loga y−3 loga x
¿ log a x
1
2 + loga y4−loga x3
¿ log a ( x
1
2 × y4 ) −loga x3
¿ loga ( x
1
2 × y4
x3 )
¿ loga ( x
1
2 × y4
x3 )
¿ log a ( x ( 1
2−3 ) × y4 )
¿ loga ( x ( −5
2 ) × y4 )
¿ loga
( y 4
x
5
2 )
Solve the equation. Give your answer as a decimal rounded to the nearest
thousandth.
9. 2x=36
Solution: 2x=36
Taking natural log both sides, we have
ln 2x=ln 36

x ln 2=ln 62
x= ( ln 62 )
ln 2
x= 2× ln 6
ln 2
x= 2× ln ( 3 ×2 )
ln 2
x= 2× ( ln 3+ ln2 )
ln 2
x=5.17
10. 2 ln ( x−5 )
ln 4 =1
Solution:
2 ln ( x−5 )
ln 4 =1
2 ln ( x−5 )
ln22 =1
2 ln ( x−5 )
2× ln 2 =1
ln ( x−5 )
ln 2 =1

ln ( x−5 )=ln 2
(x−5)=2
x=7