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