Logarithms

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LOGARITHMS

The inverse of the exponential function-- The Natural Logarithm

The graph of

y = e^x

clearly shows that it is a one to one function, hence an inverse exists. We call this inverse the natural logarithm. and write it as

y = ln x

Below is the graph of

y = e^x

and

y = ln x

By definition, they are reflections of each other across the line y = x.

$The graphs of y = e^x and y = ln(x). They are reflections of each other across the line y = x.$

Inverse Properties of Logs

Since logs and exponents cancel each other we have:

e^ln^x= x

and

ln e^x = x

Example

e^ln³ = 3 and ln(e⁵) = 5

Three Properties of Logs

Property 1: ln (uv) = ln u + ln v (The Product to Sum Rule)

Property 2: ln (u/v) = ln u - ln v (The Quotient to Difference Rule)

Property 3: ln u^r = r ln u (The Power Rule)

Example:

Expand:

ln (xy²/z)

by property 2 we have:

ln (xy²) - ln z

by property 1 we have

ln x + ln y² - ln z

By property 3 we have

ln x + 2 ln y - ln z

Exercise

Try to expand

$ ln(\sqrt{x^2 y}) $

Example

Write as a single logarithm:

4ln x - 1/2ln y + ln z

Solution

We first use property 3 to write:

ln x⁴ - ln y^1/2 + ln z

Now we use property 2:

                x⁴
        ln              + ln z
              y^1/2

Finally, we use property 3:

                x⁴ z
        ln              + ln z
              y^1/2

Exercise

Write the following as a single logarithm:

1/3 ln x + 2 ln y - 3 ln z

Example

Suppose that

ln 3 = 1.10

and that

ln 5 = 1.61

Find

ln 45

Solution

Since

45 = (5)(3²)

We have

ln 45 = ln (5)(3²) = ln 5 + ln 3²

= 1.61 + 2 ln 3

= 1.61 + 2 (1.10) = 3.81

Exponential Equations

The key to solving equations is to know how to apply the inverse of a function. When we have an exponential equation, we will use the natural logarithm to cancel the exponential.

Example

Find k if

34 = e^10k

Solution

Take the natural logarithm of both sides

ln 34 = ln e^10k

Now use the inverse property

ln 34 = 10k

Finally divide by 10

           ln 34
                        = k = 0.3526
             10

Carbon Dating

All living beings have a certain amount of radioactive carbon C₁₄ in their bodies. When the being dies the C₁₄ slowly decays with a half life of about 5600 years. Suppose a skeleton is found in Tahoe that has 42% of the original C₁₄. When did the person die?

Solution

We can use the exponential decay equation:

        y = Ce^kt

After 5600 years there is

        C/2

C₁₄ left. Substituting, we get:

        C/2 = Ce^k⁽⁵⁶⁰⁰⁾

Dividing by C,

        1/2 = e^5600k

Take ln of both sides,

        ln(0.5) = 5600k

so that

                   ln(0.5)
        k =                 = -0.000124
                   5600

The equation becomes

        y = Ce^-0.000124t

To find out when the person died, substitute

        y = 0.42C

and solve for t:

        0.42C = Ce^-0.000124t

Divide by C,

        0.42 = e^-0.000124t

Take ln of both sides,

        ln(0.42) = -0.000124t

Divide by -0.000124

                 ln(0.42)
        t =                     = 6995
               -0.000124

The person died about 7,000 years ago.

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