exponential and logarithmic equations (4.6)

21
Exponential and Logarithmic Equations (4.6) The Change of Base Formula and Using various patterns to solve equations

Upload: kasimir-moon

Post on 31-Dec-2015

16 views

Category:

Documents


1 download

DESCRIPTION

Exponential and Logarithmic Equations (4.6). The Change of Base Formula and Using various patterns to solve equations. Formative Assessment Review. I heard from several of you that going over how to find an inverse function would be helpful. Here we go:. Formative Assessment Review. - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Exponential and Logarithmic Equations (4.6)

Exponential and Logarithmic Equations (4.6)

The Change of Base Formula and

Using various patterns to solve equations

Page 2: Exponential and Logarithmic Equations (4.6)

Formative Assessment ReviewI heard from several of you that going over how to

find an inverse function would be helpful.

Here we go:

ax

xxh

xxg

xf x

4)(

)1ln()(

32)(

Page 3: Exponential and Logarithmic Equations (4.6)

Formative Assessment ReviewThe graphs:

The last one on page 3, with variables:

Also, on set four (the final page), be sure to LABEL! Without some indication of intervals, the graphs are just scribbles.

And BE NEAT! I expect to see curves that line up along asymptotes, and cross at actual intercepts.

axebxh )(

Page 4: Exponential and Logarithmic Equations (4.6)

Formative Assessment ReviewFinal hint– review your notes regularly! Even five

minutes a few times a week makes a big difference.

Page 5: Exponential and Logarithmic Equations (4.6)

Using properties from last time

Solve the equation:

ln(x+6) - ln10 = ln (x-1) - ln2

Page 6: Exponential and Logarithmic Equations (4.6)

Using properties from last time

Solve the equation:

ln(x+6) - ln10 = ln (x-1) - ln2

ln ((x+6)/10) = ln ((x-1)/2)

(x+6)/10 = (x-1)/2

2(x+6) = 10(x-1)

2x +12 = 10x -10

22 = 8x

x = 11/4

Page 7: Exponential and Logarithmic Equations (4.6)

Using properties from last time

In economics, the demand D for a product is often related to its selling price p by an equation of the form

logaD = logac - k logap

Where a and c are positive constants (why positive?) and k > 1.

1. Solve the equation for D.2. How does increasing or decreasing the selling price affect the

demand?3. Think about a and c.

Page 8: Exponential and Logarithmic Equations (4.6)

Using properties from last time

In economics, the demand D for a product is often related to its selling price p by an equation of the form

logaD = logac - k logap

Where a and c are positive constants and k > 1.

1. Solve the equation for D.

k

kaa

kaaa

aaa

pcD

pcD

pcD

pkcD

loglog

logloglog

logloglog

Page 9: Exponential and Logarithmic Equations (4.6)

Using properties from last time

In economics, the demand D for a product is often related to its selling price p by an equation of the form

logaD = logac - k logap D = c/pk

Where a and c are positive constants and k > 1.

2. How does increasing or decreasing the selling price affect the demand?

Page 10: Exponential and Logarithmic Equations (4.6)

Using properties from last time

In economics, the demand D for a product is often related to its selling price p by an equation of the form

logaD = logac - k logap

Where a and c are positive constants and k > 1.

3. Think about a and c.

What do you think a would equal? In other words, what would be a reasonable log base?

What do you think c represents? (Econ students, help here.)

Page 11: Exponential and Logarithmic Equations (4.6)

Now, a logarithmic POD

Rewrite for x:

3x = 21

What is a reasonable guess for the value of x?

Page 12: Exponential and Logarithmic Equations (4.6)

From the POD

How to solve for x?

3x = 21 x = log321

Start by taking the log of each side. log 3x = log 21Then use our third Log Property. x log 3 = log 21Then solve for x. x = (log 21)/(log 3)

x = 2.77

If I came up with these values, what have I done?

x = .845x = 7x = 1.64

Page 13: Exponential and Logarithmic Equations (4.6)

From the POD

x = log321

becomes

x = (log 21)/(log 3)

This leads to the Change of Base Formula:

logb u = (loga u)/(loga b)

= (log u)/(log b)

= (ln u)/(ln b)

Page 14: Exponential and Logarithmic Equations (4.6)

The Proof (The POD with Variables)

Set w = logb u.

Then bw = u

Log each side loga bw = loga u

w loga b = loga u

w = (loga u)/(loga b) = logb u

Page 15: Exponential and Logarithmic Equations (4.6)

Careful!

loga (u/b) = loga u - loga b

It does not equal (loga u)/(loga b).

(Major foot stomp here.)

Page 16: Exponential and Logarithmic Equations (4.6)

Use it

1. Solve for x when there are different bases.

42x+3 = 5x-2

Start by taking the log of each side. That way you set up a common base. Then you can move the exponents down.

log 42x+3 = log 5x-2

(2x+3)log 4 = (x-2)log 5

And solve.

Page 17: Exponential and Logarithmic Equations (4.6)

Use it

1. Solve for x when there are different bases.

42x+3 = 5x-2

log 42x+3 = log 5x-2

(2x+3)log 4 = (x-2)log 5

.602(2x+3) = .699(x-2)

1.204x + 1.806 = .699x – 1.398

.505x = -3.204

x = -6.345

Check it!

Page 18: Exponential and Logarithmic Equations (4.6)

Use it

2. Solve for x. (Alert! Embedded quadratic here.)

(5x - 5-x)/2 = 3

Page 19: Exponential and Logarithmic Equations (4.6)

Use it

2. Solve for x. (Alert! Embedded quadratic here.)

Does it check?

01565

5615

5

56

5

1

5

5

65

15

655

32/)55(

2

2

2

xx

xx

x

x

xx

x

xx

xx

xx

130.15log

162.6log162.6log

5162.6

162.,162.61032

1026

2

4366

016

5

5

2

x

m

mm

m

x

x

Page 20: Exponential and Logarithmic Equations (4.6)

Use it

3. If a beam of light with intensity I0 is projected vertically downward into a certain body of water, then its intensity, I(x), at a depth of x meters is

I(x) = I0e-1.4x

At what depth is its intensity half of its value at the surface?

Page 21: Exponential and Logarithmic Equations (4.6)

Use it

3. If a beam of light with intensity I0 is projected vertically downward into a certain body of water, then its intensity, I(x), at a depth of x meters is

I(x) = I0e-1.4x

So, the intensity is cut in half at about half a meter.

50.4.1

5.ln

5.ln4.1

5. 4.1

x

x

e x