8.6 natural logarithms. natural logs and “e” start by graphing y=e x the function y=e x has an...
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8.6 Natural Logarithms
Natural Logs and “e”
Start by graphing y=ex The function y=ex has an inverse called the Natural
Logarithmic Function.
Y=ln x
What do you notice about the graphs of y=ex
and y=ln x?
y=ex and y=ln x are inverses of each other!
We can use the natural log to “undo” the function y= ex (and vice versa).
All the rules still apply
• You can use your product, power and quotient rules for natural logs just like you do for regular logs
4ln2ln5
8ln4
32ln
4
2ln
5
yx lnln3 yx 3ln
Let’s try one:
Solving with base “e”
205.27 2 xe
5.177 2 xe
5.2lnln 2 xe
5.2ln2 x
5.22 xe 2. Divide both sides by 7
3. Take the natural log of both sides.
4. Simplify.
1. Subtract 2.5 from both sides
5. Divide both sides by 2
2
5.2lnx
x = 0.458 6. Calculator
Another Example: Solving with base “e”
301 xe
30lnln 1 xe
30ln1x
1. Take the natural log of both sides.
2. Simplify.
3. Subtract 1 from both sides 1)30(ln x
x = 2.401 4. Calculator
Solving a natural log problem
3)4ln( x
3)4( ex
086.16x
086.204x 2. Use a calculator
3. Simplify.
1. Rewrite in exponential form
To “undo” a natural log, we use “e”
Another Example: Solving a natural log problem
4)53ln( 2 x
42)53( ex
60.54)53( 2 x
1. Rewrite in exponential form.
2. Calculator.
3. Take the square root of each time60.5453 x
3x+5 = 7.39 or -7.39 4. Calculator
X=0.797 or -4.130 5. Simplify
Let’s try some 21)93(ln x 1.92.75
2
x
e
Let’s try some 21)93(ln x 1.92.75
2
x
e
Going back to our continuously compounding interest problems . . . A $20,000 investment appreciates 10% each
year. How long until the stock is worth $50,000?
Remember our base formula is A = Pert . . . We now have the ability to solve for t
A = $50,000 (how much the car will be worth after the depreciation)
P = $20,000 (initial value)
r = 0.10
t = time
From what we have learned, try solving for time
Going back to our continuously compounding interest problems . . . $20,000 appreciates 10% each year. How
long until the stock is worth $50,000?
A = $50,000 (how much the car will be worth after the depreciation)
P = $20,000 (initial value)
r = 0.10
t = time
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