Exponential and Logarithmic Equations (4.6)

Exponential and Logarithmic Equations (4.6)

The Change of Base Formula and

Using various patterns to solve equations

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)(

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 )(

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

minutes a few times a week makes a big difference.

Using properties from last time

Solve the equation:

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

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

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.

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

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?

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.)

Now, a logarithmic POD

Rewrite for x:

3x = 21

What is a reasonable guess for the value of x?

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

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)

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

Careful!

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

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

(Major foot stomp here.)

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.

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!

Use it

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

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

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

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?

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