Module'2'Lesson'4'
Real'World'Applications'
Exponential'and'Logarithmic'Functions'Notes'Part'1!
!
!
There!are!an!incredible!number!of!phenomena,!which!can!be!modeled!by!using!an!
exponential!or!logarithmic!function.!!Here!are!a!few:!
!
!
Interest:!
!
Simple!interest!is!when!your!bank!gives!you!a!certain!amount!of!money!yearly,!based!on!
how!much!is!in!your!account!and!what!the!interest!rate!is.!!!
!
The!formula!for!simple!interest!looks!like!this:!
I!=!Prt!!!
!
‘r’!is!the!interest!rate!(as!a!decimal!)!
P!is!the!principal!(starting)!balance!
‘t’!is!the!time!in!years.!
!
Compound!interest!is!when!your!bank!gives!you!a!certain!amount!of!money!a!number!of!
times!during!the!year.!!It!might!be!quarterly,!monthly,!or!even!daily.!!!
!
r
The!formula!for!compound!interest!looks!like!this:! A = P(1 + ) nt !
n
!
‘n’!is!the!number!of!times!interest!is!given!per!year!
!
For!example,!quarterly!interest!would!make!n!=!4!
‘A’!is!the!amount!of!money!we!have!in!the!account!after!interest!is!added!
‘P’!is!the!principal!(starting)!balance!
!
!
When!interest!is!added!to!your!account!continuously,!the!formula!changes.!!Continuous!
compound!interest!means!that!you’re!getting!interest!more!than!daily,!more!than!
hourly,!even!more!than!minutely.!!It!means!‘n’!is!getting!very,!very!large.!
!
As!‘n’!gets!infinitely!large,!the!formula!approaches!the!natural!base,!e,!and!becomes!this!
formula!for!continuously!compounded!interest:! !
!
A = Pert !
!
!
'
!
1!
Example'1:!
!
!
!
!
!
!
!
Find!the!interest!on!a!Principal!Balance!of!$10,000!over!the!course!
of!8!years!with!an!interest!rate!of!5.5%.!!Do!this!for:!
a)!
Simple!Interest!
b)!
Monthly!Compound!Interest! !
!
c)!
Continuously!Compounded!Interest!
!
a) Simple!Interest! I = Pr t !!
I = 10000(.055)8
!!
I = 4400
!
The!simple!interest!earned!in!8!years!at!a!rate!of!5.5%!would!be!$4400.!!
r
b) Monthly!Compound!Interest! A = P(1 + ) nt !
n
!
.055 12(8)
A = 10000(1+
)
12
!!
A = 15511.47
!
The!amount!15511.47!is!the!total!amount!in!the!account.!!To!calculate!
the!amount!of!interest!earned!you!would!need!to!take!the!total!amount!
in!the!account!after!8!years!and!subtract!the!principal!amount.!!
15511.47 − 10000 = 5511.47 !
!
The!interest!earned!in!8!years!at!a!rate!of!5.5%!compounded!monthly!
would!be! 5511.47 .!
!
c) Continuously!Compounded!Interest! A = Pert !
A = 10000e.055i8
!
A = 15527.07
!!
The!amount!15527.07!is!the!total!amount!in!the!account!after!8!years.!!
To!calculate!the!amount!of!interest!earned,!take!the!total!amount!in!the!
account!after!8!years!and!subtract!the!principal!amount.!!
15527.07 − 10000 = 5527.07 !!
!
The!interest!earned!in!8!years!at!a!rate!compounded!continuously!would!
be!$5527.07.!!!
!
!
!
2!
Example'2:!
!
Your!bank!account!is!sitting!on!$28,000.!!You!started!with!a!
principal!balance!of!$20,000.!!If!the!interest!rate!was!6%!
compounded!continuously,!how!long!was!your!money!in!the!
bank?!
Continuously!Compounded!Interest! A = Pert !
28000 = 20000e.06t
28000
= e.06t
20000
7
= e.06t
5
7
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ln = .06t
!!
5
7
ln
5
t=
.06
ln 7 − ln 5
t=
.06
t ≈ 5.608
!
The!money!was!in!the!bank!for!approximately!5.608!years.!!
Substitute!what!you!
know!into!the!
equation.!!Solve!for!
what!you!don’t!
know.!!
!
!
Newton’s'Law'of'Cooling:'
!
Every!good!coroner!knows!that!a!dead!body!will!eventually!reach!equilibrium!with!the!
temperature!of!its!surroundings.!!In!other!words,!the!body’s!temperature!will!go!from!
its!98.6!degree!normal!to!whatever!the!temperature!is!in!its!environment.!
!
The!function!that!this!temperature!drop!follows!is!exponential.!!It!is!given!by:!
!
T −S
!
kt = ln
T0 − S
!
!
where!‘t’!is!the!time!the!person!has!been!dead,!‘S’!is!the!surrounding!air!temperature,!!
‘T’!is!the!current!body!temperature,!and! T0 !is!the!starting!temperature!of!the!body!
(98.6)!
!
!
!
3!
Example'3:!A!coroner!finds!a!dead!body!and!checks!the!temperature.!!It!reads!80!
degrees.!!Room!temperature!is!72!degrees.!!If!the!constant,!k,!is!–
0.2,!how!many!hours!has!the!person!has!been!dead?!
!
For!the!record,!this!formula!applies!to!other!cooling!problems,!not!just!dead!bodies.!!
T −S
!
kt = ln
T0 − S
80 − 72
(−0.2)t = ln
98.6 − 72
80 − 72
ln
!!
t = 98.6 − 72
−.2
t ≈ 6.0073 hours
!
The!person!has!been!dead!for!approximately!6!hours.!!
!
HalfGlife:!
!
The!half`life!of!a!substance!is!the!amount!of!time!it!takes!for!one`half!of!the!total!
amount!of!the!substance!to!disappear!or!decay.!!!
!
x
The!formula!for!half`life!looks!like!this:!
!
⎛ 1⎞
A = Ao ⎜ ⎟ !
⎝ 2⎠
!
(Note:!!This!formula!is!simplified!from!what!you!might!find!in!a!science!book).!
!
A!=!Current!amount!of!a!substance!
Ao!=!Initial!amount!of!a!substance!
x!=!units!of!time!
!
For!example,!let’s!say!that!you!have!100!grams!of!the!238`Plutonium!isotope!(known!as!
238
Pu ).!!This!isotope!has!a!half!life!of!87!years.!
That!means!each!unit!of!x!is!87!years.!!Let’s!look!at!what!happens!to!the!element:!
!
Grams! x!
Total!years!
100!
0!
0!
50!
1!
87!
25!
2!
174!
12.5!
3!
261!
6.25!
4!
348!
3.125!
5!
435!
1.563!
6!
522!
0.781!
7!
609!
!
4!
0.391!
8!
696!
!
!
!
The!‘total!years’!column!is!found!by!taking!87!times!‘x’.!
!
In!696!years,!we!only!have!about!four`tenths!of!one!gram!of!our!original!left.!
!
Example'4:!
You!start!with!500!grams!of! 238 Pu .!!How!much!is!left!after!1000!
years?!
!
x
⎛ 1⎞
A = Ao ⎜ ⎟
⎝ 2 ⎠ !!
⎛ 1⎞
A = 500 ⎜ ⎟
⎝ 2⎠
!
'
Example'5:!
!
!
The!half!life!of!!
!is!87!
years,!you!have!to!make!x!1000!
divided!by!87!because!x!is!the!
unit!of!time.!!
1000
87
!!
A ≈ .1733
It!is!the!year!2235.!!Scientists!have!unearthed!a!rock!sample!from!
an!old!nuclear!power!plant!where!there!was!a!big!nuclear!
meltdown!in!1995.!!This!rock!sample!contains!0.2!grams!of! 238 Pu !
How!much! 238 Pu was!in!the!sample!initially?!!Round!your!answer!
to!3!decimal!places.!!
x
⎛ 1⎞
A = Ao ⎜ ⎟
⎝ 2⎠ !
⎛ 1⎞
0.2 = A0 ⎜ ⎟
⎝ 2⎠
A0 =
2235−1995
87
0.2
⎛ 1⎞
⎜⎝ ⎟⎠
2
240
87
!!
A0 ≈ 1.353
!
!
!
There!was!approximately!1.353!grams!of! 238 Pu !in!the!rock!in!1995!
at!the!time!of!the!meltdown!in!1995.!
5!
Another'way'to'do'halfGlife:!
!
Some!people!prefer!this!method!of!doing!half!life!problems.!
!
Consider!the!general!form!of!the!exponential!function:!
y = ca x !
!
We!can!use!this!equation!directly!to!solve!our!half!life!problems!because!we!can!find!a!
base!‘a’!which!incorporates!the!½!and!the!‘k’!together.!
!
‘c’!is!still!our!initial!value,!‘x’!is!still!our!time.!
!
Example'6:! The!half!life!of!an!isotope!known!as!Iodine`125!is!57.4!days.!
What!is!the!rate!of!decay!of!this!substance?!!
!
When!the!problem!wants!‘rate!of!decay’,!it!wants!the!value!for!‘a’.!!To!find!this,!let’s!
imagine!first!we!had!100!grams!initially.!!After!one!half`life,!we’d!have!50!grams!left.!!
Let’s!use!these!hypothetical!values!to!find!‘a’:!
!
y = ca x
50 = 100a57.4
1
= a57.4
2
!
1 57.4 57.4
= a
2
0.9880 = a
57.4
!
!
Hopefully!you!can!see!that!it!didn’t!matter!what!we!chose!for!our!initial!and!current!
values.!!When!you!divide!the!two!after!one!half`life,!you’ll!always!get!½.!
!
The!advantage!of!this!method!is!that!you!don’t!have!to!adjust!to!make!‘x’!the!number!of!
half`lives.!!It!can!simply!be!the!time!it!takes!for!one!half!life.!
!
Use!the!values!given!for!Iodine`125!to!answer!problems!6!and!7:!
!
!
!
!
!
!
!
!
6!
Example'7:!
!
!
!
!
If!you!start!with!100g!of!this!substance,!how!much!is!left!in!one!year?!!
Round!to!three!decimal!places.!
Since!the!half!life!is!measured!in!
days!and!in!example!6!we!solve!
for!the!“a”!using!57.4!as!the!x,!we!
have!to!use!days!as!units!for!“x.”!!
So!1!year!is!equivalent!to!365!
days.!!
y = ca x !
y = 100(.988)365
y = 1.22
!
!
!!
!
There!would!be!1.22!grams!of!Iodine`125!left!after!1!year!(365!days).!!
!
Example'8:!
!
!
Let’s!say!you!have!5g!of!Iodine`125!from!a!different!experiment.!!This!
experiment!started!with!200g.!!How!many!days!has!the!substance!been!
decaying?!
y = ca x !
5 = 200(.988)x
!
1
= .988 x
! 40
!!
1
log.988
=x
40
x ≈ 305.558
!
The!Iodine`125!has!been!decaying!for!approximately!305.557!days.!!
!
Continuous'Growth'and'Decay:!
!
Bacteria!growth!is!a!constant!event,!much!like!continuously!compounded!interest.!!
Decay!of!a!radioactive!substance,!such!as! 238 Pu !and! 14C !(Carbon`14,!used!in!
radiocarbon!dating!to!find!age!of!substances)!is!also!continuous.!!
!
The!formula!for!continuous!growth!and!decay!is:! !
A = Ao e kt !
!
‘A’!is!the!current!amount!of!material!
‘Ao’!is!the!initial!amount!of!material!
‘k’!is!a!constant,!different!for!each!process.!!!
‘k’!will!be!negative!for!decreasing!processes!and!positive!for!increasing!ones!
‘t’!is!the!time!in!years!
!
Half`life!could!also!be!calculated!as!a!continuous!decay!process.!!
!
7!
!
!
Use!the!continuous!growth!and!decay!model! A = Ao e kt !for!examples!9`11.!!
'
Example'9:!
What!is!the!rate!of!decay!if!250!grams!remain!of!a!material!that!
has!a!half`life!of!2700!years?!!Round!to!4!decimal!places.!
kt
A = Ao e !
To!find!the!rate!of!decay!of!this!problem,!250!is!
250 = 500ek⋅2700
the!current!amount!of!material!and!if!given!the!
1
= e2700 k
half!life!is!2700,!you!would!use!250!as!the!current!
2
amount!“A”!and!then!you!know!if!“t”!is!2700,!then!
1
the!initial!amount!would!have!been!500!(that!is!
ln = 2700k !!
2
double!the!current!amount).!!Plug!in!those!values!
1
and!solve!for!k.!!!
ln
2
k=
**Remember!when!give!
!the!next!step!
2700
k ≈ −.0003
you!have!to!do!is!change!that!to!logarithmic!form.!!
!
!
The!rate!of!decay!would!be!.03%.!every!year.!!
!
!
!!
Example'10:!!What!is!the!rate!of!growth!of!10!bacteria!if!after!3.5!hours,!there!are!2,000!
bacteria?!!Round!to!4!decimal!places!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
! A = Ao e kt !
Plug!in!the!values!given!and!solve!for!k.!!!
2000 = 10e 3.5 k
!
200 = e 3.5 k
**!Remember!if!“k”!is!a!negative!value!
then!it!is!a!rate!of!decay.!!If!it!is!a!positive!
ln 200 = 3.5k !
value!it!is!a!rate!of!growth.!!
ln 200
k=
3.5
k ≈ 1.5138
!
The!rate!of!growth!would!be!151.38%!every!hour!
!
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8!
Example'11:!!How!long!will!it!take!50!grams!of!a!substance!to!decay!to!10!grams!if!the!
rate!of!decay!is! k = −.345 !daily.!!!Round!to!the!nearest!tenth.!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
! A = Ao e kt !
10 = 50e−.345t
1
= e−.345t
Substitute!the!values!given!in!the!problem!and!
5
then!solve!for!the!time.!
1
ln = −.345t !
5
1
ln
5
t=
−.345
t ≈ 4.7
!
!
It!will!take!4.7!days!for!50!grams!of!a!substance!to!decay!to!10!grams!if!the!
rate!of!decay!is! k = −.345 .!!
!
9!