Exponential and Logarithmic Equations (4.6) The Change of Base Formula and Using various patterns to...
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![Page 1: Exponential and Logarithmic Equations (4.6) The Change of Base Formula and Using various patterns to solve equations.](https://reader036.fdocuments.net/reader036/viewer/2022070412/56649de85503460f94ae207a/html5/thumbnails/1.jpg)
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) The Change of Base Formula and Using various patterns to solve equations.](https://reader036.fdocuments.net/reader036/viewer/2022070412/56649de85503460f94ae207a/html5/thumbnails/2.jpg)
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)(
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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 )(
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Formative Assessment ReviewFinal hint– review your notes regularly! Even five
minutes a few times a week makes a big difference.
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Using properties from last time
Solve the equation:
ln(x+6) - ln10 = ln (x-1) - ln2
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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
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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.
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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
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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) The Change of Base Formula and Using various patterns to solve equations.](https://reader036.fdocuments.net/reader036/viewer/2022070412/56649de85503460f94ae207a/html5/thumbnails/10.jpg)
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) The Change of Base Formula and Using various patterns to solve equations.](https://reader036.fdocuments.net/reader036/viewer/2022070412/56649de85503460f94ae207a/html5/thumbnails/11.jpg)
Now, a logarithmic POD
Rewrite for x:
3x = 21
What is a reasonable guess for the value of x?
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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
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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)
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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
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Careful!
loga (u/b) = loga u - loga b
It does not equal (loga u)/(loga b).
(Major foot stomp here.)
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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.
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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!
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Use it
2. Solve for x. (Alert! Embedded quadratic here.)
(5x - 5-x)/2 = 3
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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
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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?
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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