Section 4.6 Modeling with Exponential and Logarithmic Functions
Exponential and Logarithmic Equations (4.6) The Change of Base Formula and Using various patterns to...
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Transcript of Exponential and Logarithmic Equations (4.6) The Change of Base Formula and Using various patterns to...
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