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Exponential Growth

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Exponential Growth Laws of Exponents and Geometric Patterns * * Laws of Exponents Consider the expression 2x4 With this we can get the Laws of exponents Multiplying ... – PowerPoint PPT presentation

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Title: Exponential Growth


1
Exponential Growth
  • Laws of Exponents and Geometric Patterns

2
Laws of Exponents
The 2 is called the coefficient and the x4 is
called the power. x is the base and 4 is the
exponent
Consider the expression 2x4
2 (x)(x)(x)(x)
With this we can get the Laws of exponents
From this we see that when multiplying powers
with the same base, we keep the base and add the
exponents.
Multiplying Powers with the same base
(a5)(a2) (aaaaa)(aa)
(a5)(a2) aaaaaaa (a5)(a2) a7
3
Laws of Exponents
Recall that 2x4 2(x)(x)(x)(x)
With this we can get the Laws of exponents
From this we see that when multiplying powers
with the same base, we keep the base and add the
exponents.
Multiplying Powers with the same base
From this we see that when dividing powers with
the same base, we keep the base and subtract the
exponents.
Dividing Powers with the same base
4
Laws of Exponents
Recall that 2x4 2(x)(x)(x)(x)
With this we can get the Laws of exponents
From this we see that when multiplying powers
with the same base, we keep the base and add the
exponents.
Multiplying Powers with the same base
From this we see that when dividing powers with
the same base, we keep the base and subtract the
exponents.
Dividing Powers with the same base
When we have a power of a power, we keep the base
and multiply the exponents.
Power of a Power
5
Using the Laws of Exponents
Simplify the following expressions
  • Answers
  • w13 b) 40d8
  • c) 5s2 d) 211
  • e) s8t8 f) 23

6
Zero and Negative Exponents
We also know that
23 2x2x2 8
When dividing powers with the same base, we keep
the base and subtract the exponents. So
So 20 must equal 1. In fact, ANYTHING to the zero
equals 1.
b0 1
7
Zero and Negative Exponents
What does equal?
We also know that
32 3x3 9 33 3x3x327
When dividing powers with the same base, we keep
the base and subtract the exponents. So
So 3-1 must equal 1/3.
A negative exponent means you flip the base and
keep the positive exponent
8
Zero and Negative Exponents
Ex1. Simplify the following leaving no negative
exponents. a)
b)
c)
Ex2. Evaluate (which means get the value of the
expression)
9
Zero and Negative Exponents

Ex 3. Evaluate each expression
10
Changing the base
The laws of exponents only work when theres a
common base. Sometimes we can change the base of
a power to make it common to another.
All of these bases can be written as a base of 2.
This doesnt look like simplifying.
Ok maybe it does
Yep, thats simpler. It is now expressed as a
single base.
11
Changing the base
This expression has some nasty values! You are
NOT required to know 81-5 or (1/3)7. BUT
Power of a Power keep the base and multiply the
exponents
All of these bases can be written as a base of 3.
Multiplying powers with the same base keep the
base and add the exponents
Dividing powers with the same base keep the
base and subtract the exponents
3
12
Solving Exponential Equations
When solving exponential equations (right now) we
will look to get a common base.
If the bases are equal, then the exponents must
be equal
Ex 1. Solve for x
or Find the Root of the equation
We must recognize that these bases can be written
using base 3.
Power of a power keep the base and multiply the
exponents
13
Solving Exponential Equations
Ex 2. Solve for x.
Recognize the base of 2
Power of a power keep the base and multiply the
exponents
If the bases are equal, then the exponents must
be equal
14
Solving Exponential Equations
Ex 3. Find the root(s) of the equation below.
ISOLATE THE POWER!!
Base of 2
15
Solving Exponential Equations
Your turn Find the root(s) of the equation
below.
16
Solving Exponential Equations
Your turn Find the root(s) of the equation
below.
17
Solving Exponential Equations
Your turn Find the root(s) of the equation
below.
18
Patterns (again)
We have seen that a pattern can be represented as
an equation.
Linear Pattern
CD is on D1
Quadratic Pattern
CD is on D2
But what about a pattern like this?
3,6,12,24,48
19
Geometric Patterns
We can quickly see that there is no common
difference for this pattern.
We can divide the terms to find a COMMON RATIO
3, 6, 12, 24, 48
3 6 12 24
Patterns with a COMMON RATIO are called Geometric
Patterns
So we can express this pattern as
3, 6, 12, 24, 48
3 3 x 2 3 x 2 x 2 3 x 2 x 2 x 2 3 x 2 x 2 x 2 x 2
2 2 2 2
Or
3 3 x 21 3 x 22 3 x 23 3 x 24
x 20
Remember 20 1
20
Geometric Patterns
3, 6, 12, 24, 48
3 3 x 21 3 x 22 3 x 23 3 x 24
x 20
We see from this that this geometric pattern can
be represented by the equation
Lets check If n 4
The 2 is the common ratio of the pattern and is
the base in the equation.
The 3 is the first term of the pattern and is the
coefficient in the equation.
21
Geometric Patterns
In general, a geometric pattern can be written
using the equation
Where t1 is the first term of the pattern (when
n1) and where r is the common ratio.
The pattern 5, 10, 20, 40, 80, can be
represented by the equation
We can check by plugging in n 5
22
Geometric Patterns
Find the 10th term in each pattern
  • 100, 50, 25, 12.5,
  • 0.25, 1,4, 16,
  • c) 5, 8, 11, 14,
  • d) 1, -2, 4, -8, 16,
  1. CR ½ t1 100

23
Geometric Patterns
Find the 10th term in each pattern
  • 100, 50, 25, 12.5,
  • 0.25, 1,4, 16,
  • c) 5, 8, 11, 14,
  • d) 1, -2, 4, -8, 16,

b) CR 4 t1 0.25
24
Geometric Patterns
Find the 10th term in each pattern
  • 100, 50, 25, 12.5,
  • 0.25, 1,4, 16,
  • c) 5, 8, 11, 14,
  • d) 1, -2, 4, -8, 16,

c) CD 3 t1 5
25
Geometric Patterns
Find the 10th term in each pattern
  • 100, 50, 25, 12.5,
  • 0.25, 1,4, 16,
  • c) 5, 8, 11, 14,
  • d) 1, -2, 4, -8, 16,

d) CR -2 t1 1
26
Geometric Patterns
For the pattern below, which term has a value of
768?
6,12,24,48,
The 8th term has a value of 768.
27
May I Have Another Word (Problem)
A great many things in nature grow exponentially.
Each of these situations can be modeled with a
geometric pattern and thus an exponential
equation.
Ex. A certain type of bacteria doubles every six
hours. The experiment begins with 1 bacteria.
Yet another 6 hours later
Another 6 hours later
6 hours later
28
May I Have Another Word (Problem)
Ex. A certain type of bacteria doubles every six
hours. The experiment begins with 1 bacteria.
After a short while
29
May I Have Another Word (Problem)
We can set up a pattern to show the number of
bacteria.
Ex. A certain type of bacteria doubles every six
hours. The experiment begins with 1 bacteria.
Number of hours since the experiment began 0 6 12 18
Number of bacteria present 1 2 4 8
We see theres a CR of 2 in the pattern.
However, this pattern involves n values that do
not increase by 1. They increase by 6.
30
May I Have Another Word (Problem)
We can set up a pattern to show the number of
bacteria.
Ex. A certain type of bacteria doubles every six
hours. The experiment begins with 1 bacteria.
Number of hours since the experiment began 0 6 12 18
Number of bacteria present 1 2 4 8
We see theres a CR of 2 in the pattern.
When this happens, the equation must change.
However, this pattern involves n values that do
not increase by 1. They increase by 6.
Notice as well that we do not know the t1 value.
We know t0 and t6 instead.
We cannot use
31
May I Have Another Word (Problem)
We can set up a pattern to show the number of
bacteria.
Ex. A certain type of bacteria doubles every six
hours. The experiment begins with 1 bacteria.
Number of hours since the experiment began 0 6 12 18
Number of bacteria present 1 2 4 8
Well need to use the equation
Where y is the amount at time x and A0 is the
original amount and the period is the amount by
which the x values increase.
So we get the function
32
May I Have Another Word (Problem)
Ex. A certain type of bacteria doubles every six
hours. The experiment begins with 1 bacteria.
How many bacteria are present after 50 hours?
Just from the question we can see that r 2
(doubles) and t0 1so we can get the equation
Now let x 50 and solve for y.
There would be 322 bacteria at t 50 hours.
33
May I Have Another Word (Problem)
Ex. A certain type of bacteria doubles every six
hours. The experiment begins with 1 bacteria.
When will there be 128 bacteria present?
Now let y 128 and solve for x.
At t42 hours there will be 128 bacteria.
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