Unit 5 (Part 1): Exponential (power) number vs logarithmic ... · MCU504/Unit 5: Exponential Number...

1 MCU504/Unit 5: Exponential Number Vs Logarithmic numbers, Financial Mathematics -Mr. Randimbiarison DMHS

Unit 5 (Part 1): Exponential (power) number vs logarithmic number

A function often is expressed in the form of y = f(x). This notation materializes the relation between the

dependent variable y and the independent variable x, i.e., y is said to be dependent of x, or y is a function of x:

when x changes in value, y changes accordingly.

An exponential number or generally expressed as an exponential function is represented by an equation in its

basic form:

𝒇(𝒙) = 𝒂𝒄𝒙 or 𝒚 = 𝒂𝒄𝒙 or 𝒚 = 𝒂 × 𝒄 × 𝒄 × 𝒄 × 𝒄 × 𝒄 × … … . .× 𝒄

6

y is the power value or the exponential value

a is the initial amount (quantity) or initial value

c is the base of the exponential function (number)

x is the exponent or the number of times the base c is multiplied by itself

Remember that for any value c raised to the power of zero, c is always equal to 1:

𝒄𝟎 = 1 (Replace c by 123,456,789 on your calculator then raise it to the power of 0. What did you find?)


If c = 0, c raised to the power of zero is UNDEFINED.

𝟎𝟎 = UNDEFINED (𝟎𝟎 this will throw an error on your calculator, such as ERROR2,…)

Few case studies of operations and graphing of exponential functions:

Let’s experiment with few values of a, c, and x. Calculate the value of y, such that 𝒚 = 𝒂𝒄𝒙 in the following

scenarios:

Case where the base c = 1:

1) If c = 1, a = 1, x = 0 → y = 𝟏(𝟏)𝟎 = 1x1=1 → since any number to the power of zero is equal to 1

2) If c = 1, a = 1000, x = 1 → y = 𝟏𝟎𝟎𝟎(𝟏)𝟏 = 1000x1=1000 → since any number to the power of 1 is equal

to that same number

3) If c = 1, a = 1000, x = 1

2 or x = 0.5 → y = 𝟏𝟎𝟎𝟎(𝟏)𝟎.𝟓 is the same as y=1000x√1=1000

→ since 1 raised to the power of any number 1 is always equal to 1

Note that, raising a number c to the power of an exponent of the form 1

𝑛 , where n is a positive integer value, is

the same as calculating the n-th root of that number c.


Case where the base c >1 and the exponent x ≥ 0

Example: if c = 1.06 and x ≥ 0 and a = 1:→ 𝒚 = (𝟏)(𝟏. 𝟎𝟔)𝒙

X=0 𝑦 = (1)(1.06)0= 1(1) = 𝟏

• any number to the power of zero is equal to 1

X=1 𝑦 = (1)(1.06)1= 1(1) = 𝟏.06

• Same as 𝑦 =(1)(1.06)0 × 1.06

X=2 𝑦 = (1)(1.06)2= 1(1.1236) = 1.1236

• Same as 𝑦 =(1)(1.06)1 × 1.06

X=3 𝑦 = (1)(1.06)3= 1(1.191016) = 1.191016

• Same as 𝑦 =(1)(1.06)2 × 1.06

….. ….. …..

X=20 𝑦 = (1)(1.06)20= 1(1.191016) = 3.207….

• Same as 𝑦 =(1)(1.06)19 × 1.06

x→ ∞ 𝑦 = (1)(1.06)𝑥→ ∞

• y goes to infinity as x

goes to ∞


Case where the base c ≥ 1 and the exponent x is such that 0 < x < 1:

Example: if c =2 and 0 < x <1 and a = 1 where 𝑦 = (1)(2)𝑥

X = 1

2

𝑦 = (1)21

2= √22

=

√2=1.4142…

• 2nd -root of 2 or commonly known as square root of 2

X = 13

𝑦 = (1)21

3=√23

= 1.25992…

• 3rd-root of 2 or commonly known as cube root of 2

• on your calculator you can raise 2 to the power of (1÷3) if you don’t have the key

√3

X = 1

4

𝑦 = (1)21

4=√24

= 1.189207…

• 4th -root of 2

• on your calculator you can raise 2 to the power of (1÷4) to obtain the result on the left Enter: 1 x (2 ^ (1÷4)) press “=

….. ….. ……

X= 1

3000

𝑦 = (1)21

3000= √23000

= 1.0002310…

• X = 1𝑛 as n gets very large,

y approaches closer to 1

X = 1

𝑛

n→ ∞: x is closer to 0

𝑦 = (1)21

𝑛= √2𝑛

(in root form) y → 1

• nth -root of 2, for any n value

• when n goes to infinity, 1

𝑛 𝑤ill eventually be near

zero, which makes y = 1, since (2)0 →1x1 = 1


Case where the base 0 < c < 1 and the exponent x ≥ 0:

Example: if c = 0.8 and x ≥ 0 and a=100 where 𝑦 = (100)(0.8)𝑥

X=0 𝑦 = (100)(0.8)0= 𝟏00

• any number to the power of zero is equal to 1

X=1 𝑦 = (100)(0.8)1= 𝟖𝟎

• any number to the power of 1 is equal to that same number

X=2 𝑦 = (100)(0.8)2= 𝟔𝟒

• the power value of 0.8 is decreasing noticeably by a factor of 0.8 from the previous power value: 80 x 0.8 = 64

…. ….. ……

X=30 𝑦 = (100)(0.8)30= 𝟎. 𝟏𝟐𝟑𝟕𝟗𝟒

• The power value 0.8 decreases closer to zero as the exponent grows larger

x→ ∞

𝑦 = (100)(0.8)𝑥= closer to zero, when x goes to infinity

• y decreases to

near 0 when

x → ∞


Examples of operations with exponential equations

a) The initial quantity a, the base c, and the exponent x are all known, but y is unknown:

𝐸𝑥: 𝑦 = 2600(0.8)𝑥. Calculate y when x = 9

Answer: 𝑦 = 2600(0.8)9 By entering 𝟐𝟔𝟎𝟎(𝟎. 𝟖)𝟗 on your calculator→ y = 348.97

b) The value of y, the base c, and the exponent x are known, but the initial quantity a is unknown:

𝑦 = 𝑎(1.08)𝑥. Calculate a when y = 400, c = 1.08, and x = 5

→ 400 = 𝑎(1.08)5

→by dividing by (1.08)5 on both sides → 400

(1.08)5=

𝑎(1.08)5

(1.08)5 →

400

(1.08)5= 𝑎

→ Using your calculator directly→ 400 ÷ (1.08^5) = 272. 2332788→ a = 272.2332788

Check: y? =400 → 𝑦 = 𝟐𝟕𝟐. 𝟐𝟑𝟑𝟐𝟕𝟖𝟖(1.08)5→ y = 400

To check on your calculator, enter 𝟐𝟕𝟐. 𝟐𝟑𝟑𝟐𝟕𝟖𝟖 × (𝟏. 𝟎𝟖)^𝟓

c) The value of y, the initial quantity a, and the exponent x are known, but the base c is unknown:

𝑦 = 𝑎(𝑐)𝑥. Calculate c when y = 1200, a = 150 and x = 5

𝟏𝟐𝟎𝟎 = 𝟏𝟓𝟎(𝒄)𝟓. Calculate the base c.

First, to isolate(𝒄)𝟓, divide both sides by 150 → 1200

150=

150 (𝑐)5

150 →

1200

150= (𝑐)5

To obtain the value of the base c, calculate the 5th root of 1200

150, or raise

1200

150 to the power of

1

5

On your calculator enter the following: (1200 ÷ 150) ^ (1÷5) = answer → c = 1.515716567


d) The y-value y, the initial quantity a, and the base c are known, but the exponent x is unknown:

𝑦 = 𝑎(𝑐)𝑥. Calculate x when y = 1200, a = 150 and c = 1.515716567

→ 1200 = 150(1.515716567)𝑥

→ by dividing by 150 on both sides 1200

150= (1.515716567)𝑥

→ When calculating the exponent x of an exponential function, from now on, say good bye to the trial and error

approach. Use the logarithmic function such that when 𝑦 = 𝑎(𝑐)𝑥 then 𝑥 =

log (𝑦

𝑎)

log 𝑐

→ 𝑥 =log (

1200

150)

𝑙𝑜𝑔1.515716567 . To calculate this expression directly on your calculator, enter the following:

𝐥𝐨𝐠 (𝟏𝟐𝟎𝟎 ÷ 𝟏𝟓𝟎) ÷ 𝒍𝒐𝒈 𝟏. 𝟓𝟏𝟓𝟕𝟏𝟔𝟓𝟔𝟕 → answer: 4.999999996 or n = 5

or enter 𝐥𝐨𝐠 (𝟏𝟐𝟎𝟎 ÷ 𝟏𝟓𝟎) ÷( 𝒍𝒐𝒈 𝟏. 𝟓𝟏𝟓𝟕𝟏𝟔𝟓𝟔𝟕)

Note: In general, if 𝒚 = 𝒂. 𝒄𝒙 or 𝒚

𝒂= 𝒄𝒙 then 𝒙 =

𝐥𝐨𝐠 (𝒚

𝒂)

𝐥𝐨𝐠 𝒄. In other words, if

𝒚

𝒂 is the

exponential value of 𝒄𝒙, then x, the exponent, is the logarithm of 𝒚

𝒂 in base c.

If a = 1 → 𝒚 = 𝒄𝒙→ 𝒙 =

𝐥𝐨𝐠 𝒚

𝐥𝐨𝐠 𝒄 . In other words, if y is the exponential value of 𝒄𝒙, then x, the

exponent, is the logarithm of y in base c.


Logarithmic Function

The Logarithmic function is the inverse of the exponential function. The graph of a

logarithmic function would be symmetrical to the exponential function over the line y = x.

Example:

The exponential and logarithm functions are inverse of each other.


A few properties or rules to retain with logarithm numbers:

When the base of a logarithm number is not specified, it is by default equal to a

base 10.

Solve the following: a) 𝑙𝑜𝑔28𝑥=40 b) 𝑙𝑜𝑔39𝑥=27 c) 𝑙𝑜𝑔3(9𝑥 − 3) = 𝑙𝑜𝑔3( 4 − 𝑥)


a) Solve for y: 𝒍𝒐𝒈𝟐 𝟏𝟐𝒚=1.77815125

Answer:

1st Method: Using the product rule: 𝑙𝑜𝑔2 12𝑦 = 𝑙𝑜𝑔2 12 + 𝑙𝑜𝑔2 𝑦 = 1.77815125

→ 𝑙𝑜𝑔2 𝑦 = 1.77815125 - 𝑙𝑜𝑔2 12 → 𝑙𝑜𝑔2 𝑦 = 1.77815125 − 𝑙𝑜𝑔2 12

1

On your calculator enter the following: 2 ^ (1.77815125 – (log12 ÷ log 2)) answer→ y = 0.285821975

Check:𝑙𝑜𝑔2 12(𝟎. 𝟐𝟖𝟓𝟖𝟐𝟏𝟗𝟕𝟓) 𝒍𝒐𝒈𝟐 𝟏𝟐𝒚=1.77815125→ 𝑙𝑜𝑔( 12(𝟎. 𝟐𝟖𝟓𝟖𝟐𝟏𝟗𝟕𝟓))÷ log 2 = 1.77815125

Better Approach:

2nd Method: → 𝑙𝑜𝑔2 12𝑦 = 1.77815125 → 12y = (𝟐)(1.77815125) → =

12y

12 = =

(𝟐)(1.77815125)

12 → y = 0.285821975

b) Solve for y: 𝒍𝒐𝒈𝟏𝟐𝒚=1.77815125 (if the base is not mentioned the default is 10: i.e., you don’t need to enter 𝑙𝑜𝑔10 in your calculator)

Answer:

1st method: using the product rule: 𝑙𝑜𝑔12𝑦 = 𝑙𝑜𝑔 12 + 𝑙𝑜𝑔 𝑦 = 1.77815125

→ 𝑙𝑜𝑔 𝑦 = 1.77815125 - 𝑙𝑜𝑔12 → 𝑙𝑜𝑔 𝑦 = 1.77815125 − 𝑙𝑜𝑔12

1

𝒚 = (𝟏𝟎)(1.77815125−log 12) . 𝒚 = (𝟏𝟎)(1.77815125−log 12) . On your calculator, enter 10 ^ (1.77815125 – (log12 ÷ log 10))

or simply enter 10 ^ (1.77815125 – log12) → answer→ y = 4.999999996 or y = 5

Check: log 12x5 = log 60? = 1.77815125 (Check)

2nd Method: → 𝑙𝑜𝑔 12𝑦 = 1.77815125 →because it’s in base 10, we can write →𝟏𝟐𝑦 = (𝟏𝟎)(1.77815125) → 𝑦 = (𝟏𝟎)

(1.77815125)

𝟏𝟐 = 5

(Better approach)

On your calculator, enter (10 ^ 1.77815125) ÷ 12


Exponential Growth (Growth is synonym of rise, increase, appreciation, augmentation, etc.)

Exponential growth models apply to any situation where the growth is proportional to the current size of the quantity of interest.

Exponential growth models are often used for real-world situations like interest earned on an investment, human or animal population, bacterial culture growth, etc.

The general exponential growth model is

→ y increases in value as t increases

where a is the initial amount or number, r is the growth rate (for example, r =2% growth rate means r=0.02 in decimal form), and t (the independent variable, same as x) is the time elapsed.

𝒚 = 𝒂(𝟏 + 𝒓)𝒕


Example 1: Angelstown has a population of 32,000. It grows with a 5% annual rate. This situation can be modeled by the

equation:

𝒚 = 𝟑𝟐𝟎𝟎𝟎(𝟏 + 𝟎. 𝟎𝟓)𝒕 (Convert the rate to decimal value: Rate 5% = 5

100 = 5÷100 = 0.05)

or 𝒚 = 𝟑𝟐𝟎𝟎𝟎(𝟏. 𝟎𝟓)𝒕, with t the number of years.

Question: After how many years (t) will the population of Angeltown reaches 140,000?

Answer: y = 140,000 → 140,000 = 32000(1.05)𝑡 → 𝑡 =log (

140 000

32000)

𝑙𝑜𝑔 1.05 → t = 30.25 years

On your calculator, enter the following: log (140000÷32000) ÷ log1.05


Sometimes, you may be given a doubling or tripling rate rather than a growth rate in percent (%).

For example, if you are told that the number of cells in a bacterial culture doubles every hour, then the equation to model the

situation would be:

Doubling rate: 𝒚 = 𝒂(𝟐)𝒕 or tripling rate: 𝒚 = 𝒂(𝟑)𝒕 with t is the number of hours elapsed, a is the initial quantity.

Example 2: Suppose a culture of 100 bacteria is put into a petri dish and the culture doubles in size every hour. Predict the number of bacteria that will be in the dish after 12 hours. Solution:

𝒚 = 𝒇(𝒕) = 𝒂(𝟐)𝒕 where a = 100 bacteria, the base a = 2, t is the time in hours

𝒚 = 𝟏𝟎𝟎(𝟐)𝟏𝟐

Enter the following on your calculator: 100 x (2^12) → answer → y = 409,600 bacteria


Example 3:

Suppose a culture of 20 bacteria is put into a petri dish and the culture triples in size every hour. After how many hours does it attain a number of 1,200,000 bacteria. Solution:

𝒚 = 𝒇(𝒕) = 𝒂(𝟑)𝒕 where a = 20 and y = 1,200,00 bacteria after t hours. t?

𝟏, 𝟐𝟎𝟎, 𝟎𝟎𝟎 = 𝟐𝟎(𝟑)𝒕

→ 1,200,000 = 20(3)𝑡

→ by dividing by 20 on both sides 1,200,000

20= (3)𝑡

→ Then calculate 𝑡 =log (

1,200,000

20)

𝑙𝑜𝑔 3

→ Enter the following on your calculator: 𝐥𝐨𝐠 (𝟏 𝟐𝟎𝟎 𝟎𝟎𝟎 ÷ 𝟐𝟎) ÷ 𝒍𝒐𝒈 𝟑 → answer: t = 10 hours


Exponential Decay (Decay is synonym of decline, decrease, depreciation, return, etc.)

Exponential decay models apply to any situation where the decay is proportional to the current size of the quantity of interest. In other words, the value of the initial quantity decreases over time.

Exponential decay models are often used for real-world situations such as depreciation (or decrease) of the price of a car, or a decrease of animal population of endangered species, etc.

The general exponential decay model is

→ y decreases in value as t increases

where a is the initial amount(quantity) or initial value, r is the rate of decay (for example, r =0.5% decay rate r=0.005 in decimal form),

and t (or x) is the time elapsed.

Note that the Rate of 0.5% = 0.5

100 = 0.5÷100 = 0.005

Example 1:

You bought a second hand 2018 Honda Civic for $19,690 from a Honda dealer. Your car will depreciate in value by a rate of 11% annually. What will be the resale price value of your car in 10 years? Solution:

𝒚 = 𝒂(𝟏 − 𝒓)𝒕 where a = 19690, r = 11% or r = 0.11, and t = 10 years

𝒚 = 𝟏𝟗𝟔𝟗𝟎(𝟏 − 𝟎. 𝟏𝟏)𝟏𝟎 → y = 6139.68

On your calculator, enter: 19690 x (1 – 0.11) ^10 → answer = 6139.680654

Conclusion: Your car is worth $ 6139.68 in 10 years

𝒚 = 𝒂(𝟏 − 𝒓)𝒕


Example 2:

Sandra bought a second hand 2017 Subaru Forester for $22,500. She wants to know when will her car worth about $10,000 if the depreciation price rate is 7% every year.

Solution:

𝒚 = 𝒂(𝟏 − 𝒓)𝒕 where a = 22,500, r = 7% or r = 0.07, y = 10,000: t =?

𝟏𝟎𝟎𝟎𝟎 = 𝟐𝟐𝟓𝟎𝟎(𝟏 − 𝟎. 𝟎𝟕)𝒕 → 𝑡 =log (

10000

22500)

𝑙𝑜𝑔 (1−0.07)

→ Enter the following on your calculator: log (10000÷22500) ÷ 𝐥𝐨𝐠 (𝟏 − 𝟎. 𝟎𝟕)

→ 𝑡 = 11.17

Conclusion: Approximately in 11 years your car is worth $10,000


Example 3:

Newfoundland cod stocks suffer serious and surprising decline. About 12,000 tonnes of cod were harvested in 2018. The cod population continues to decline at an alarming average rate of 6% annually. Approximately, how many tonnes of cod fish were harvested in 2008 knowing that this issue has been recorded in the past three decades?

Solution:

𝒚 = 𝒂(𝟏 − 𝒓)𝒕 where y = 12,000, r = 6% or r = 0.06, t =2018 – 2008 = 10 years: a =?

𝟏𝟐𝟎𝟎𝟎 = 𝒂(𝟏 − 𝟎. 𝟎𝟔)𝟏𝟎 , isolate 𝒂 by dividing by (𝟏 − 𝟎. 𝟎𝟔)𝟏𝟎on both sides

12000

(𝟏−𝟎.𝟎𝟔)𝟏𝟎=

𝒂(𝟏−𝟎.𝟎𝟔)𝟏𝟎

(𝟏−𝟎.𝟎𝟔)𝟏𝟎 →

12000

(𝟏−𝟎.𝟎𝟔)𝟏𝟎= 𝑎

→ Enter the following on your calculator: 12000÷ ((1 - 0.06) ^10) → 𝑎 = 22,279.35

Conclusion: Approximately 22,280 tonnes of cod fish were harvested in 2008


Very Important note: When will the rule in the form of 𝒚 = 𝒂(𝒄)𝒃𝒕 𝒐𝒓 𝒚 = 𝒂(𝒄)𝒃𝒙 be used?

Example 1: Suppose you would like to study the growth of a culture of 100 bacteria. We saw how you would write the rule

when the culture doubles in size every hour, 𝒚 = 𝒂(𝒄)𝒕, where b=1, since the growth is on every unit of time.

Question: Predict the number of bacteria that will be in the petri dish after 12 hours in the following scenarios:

The bacteria: a) doubles in size every half hour? b) doubles in size every two hours? c) double in size every two days?

Some people may “intuitively” write the following rules for each scenario:

a) 𝒚 = 𝟏𝟎𝟎(𝟐)𝟎.𝟓𝒕 , t = # hours

b) 𝒚 = 𝟏𝟎𝟎(𝟐)𝟐𝒕 , t = #hours

c) 𝒚 = 𝟏𝟎𝟎(𝟐)𝟐𝒕 , t = #days The rules a), b), c) above are wrong.


Solution: Although it is possible to write an exponential rule by intuition, the safest way to find it is by using the following steps:

1) Corner the unit of time t in question. Use the time unit that describes how often the base c is multiplied by itself. 2) Use a small table of value and pay attention to the pattern for each value of t.

3) Calculate the value of the coefficient b in the general equation 𝒚 = 𝒂(𝒄)𝒃𝒕

a) The bacteria doubles in size every half hour? Initial amount a = 100 bacteria

t = # hours 𝒚 = 𝒂(𝒄)𝒃𝒕 y You can continue, if you want, and enter more

values of t in the table. The two entries are enough to determine the value of b.

0 𝒚 = 𝟏𝟎𝟎(𝟐)𝒃.𝟎 100

0.5 𝒚 = 𝟏𝟎𝟎(𝟐)𝒃.(𝟎.𝟓) 200 (doubled every half hour)

When t = 0.5 y=200 → 200 = 100(2)𝑏(0.5)or 200 = 100(2)0.5𝑏

Because, I want to isolate the term with the unknown value b, divide both sides by 100 → 𝟐𝟎𝟎

𝟏𝟎𝟎=

𝟏𝟎𝟎(𝟐)𝟎.𝟓𝒃

𝟏𝟎𝟎

→ 2 = (2)0.5𝑏

Using logarithm number → 0.5b = 𝑙𝑜𝑔 2

𝑙𝑜𝑔 2 = 1 → 0.5b = 1 → b = 2

Therefore, the rule of scenario a) is 𝒚 = 𝟏𝟎𝟎(𝟐)𝟐𝒕

Conclusion: In 12 hours the number of bacteria will be 𝒚 = 𝟏𝟎𝟎(𝟐)𝟐(𝟏𝟐)= 𝒚 = 𝟏𝟎𝟎(𝟐)𝟐𝟒 → y = 1,677,721,600 (approximately, over a billion and a half bacteria)


b) The bacteria doubles in size every two hours? Initial amount a = 100 bacteria

t = # hours 𝒚 = 𝒂(𝒄)𝒃.𝒕 y

0 𝒚 = 𝟏𝟎𝟎(𝟐)𝒃.𝟎 100

2 𝒚 = 𝟏𝟎𝟎(𝟐)𝒃.𝟐 200 (doubled every two hours)

When t = 2 y=200 → 200 = 100(2)𝑏(2)or 200 = 100(2)2𝑏

To determine b, isolate the term with the unknown value b, divide by 100 on both sides → 𝟐𝟎𝟎

𝟏𝟎𝟎=

𝟏𝟎𝟎(𝟐)𝟐𝒃

𝟏𝟎𝟎

→ 2 = (2)2𝑏

Using logarithm number → 2b = 𝑙𝑜𝑔 2

𝑙𝑜𝑔 2 = 1 → 2b = 1 → b =

1

2

Therefore, the rule of scenario b) is 𝒚 = 𝟏𝟎𝟎(𝟐)𝟏

𝟐𝒕 or 𝒚 = 𝟏𝟎𝟎(𝟐)

𝒕

𝟐

Conclusion: In 12 hours the number of bacteria will be 𝒚 = 𝟏𝟎𝟎(𝟐)𝟏𝟐

𝟐 = 𝒚 = 𝟏𝟎𝟎(𝟐)𝟔 → y = 6400


c) The bacteria doubles in size every two days? Initial amount a = 100 bacteria

t = # hours 𝒚 = 𝒂(𝒄)𝒃.𝒕 y 2 days = 48 hours

0 𝒚 = 𝟏𝟎𝟎(𝟐)𝒃.𝟎 100

48 𝒚 = 𝟏𝟎𝟎(𝟐)𝒃.𝟒𝟖 200 (doubled every two hours)

When t = 48, y=200 → 200 = 100(2)𝑏(48)or 200 = 100(2)48𝑏

To determine b, isolate the term with the unknown value b, divide by 100 on both sides → 𝟐𝟎𝟎

𝟏𝟎𝟎=

𝟏𝟎𝟎(𝟐)𝟒𝟖𝒃

𝟏𝟎𝟎

→ 2 = (2)48𝑏

Using logarithm number → 48b = 𝑙𝑜𝑔 2

𝑙𝑜𝑔 2 = 1 → 48b = 1 → b =

1

48

Therefore, the rule of scenario c) is 𝒚 = 𝟏𝟎𝟎(𝟐)𝟏

𝟒𝟖𝒕 or 𝒚 = 𝟏𝟎𝟎(𝟐)

𝒕

𝟒𝟖 On your calculator, enter: 100 x (2^ (1÷4))

Conclusion: In 12 hours the number of bacteria will be 𝒚 = 𝟏𝟎𝟎(𝟐)𝟏𝟐

𝟒𝟖 = 𝒚 = 𝟏𝟎𝟎(𝟐)𝟏

𝟒 → approximately y = 118 bacteria

*Another way of solving scenario c) would be to choose the time unit as t = # days:

t = # days 𝒚 = 𝒂(𝒄)𝒃.𝒕 y When t = 2, y=200 → 200 = 100(2)𝑏.2

or 200 = 100(2)2𝑏 → 2b = 𝑙𝑜𝑔 2

𝑙𝑜𝑔 2 → b =

𝟏

𝟐

The rule is 𝒚 = 𝟏𝟎𝟎(𝟐)𝟏

𝟐𝒕

t=12 hours = 0.5 day → 𝒚 = 𝟏𝟎𝟎(𝟐)𝟏

𝟐(𝟎.𝟓)or 𝒚 = 𝟏𝟎𝟎(𝟐)

𝟏

𝟒

0 𝒚 = 𝟏𝟎𝟎(𝟐)𝒃.𝟎 100

2 𝒚 = 𝟏𝟎𝟎(𝟐)𝒃.𝟐 200 (doubled every two days)

The answer will be the same: In 12 hours or 0.5 days the bacteria grows up to about 118.

Date post:	12-Aug-2020
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