Log Bases and Exponentials Log Equations

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Log Bases and Log equations

The Common Logarithm

In chemistry, base 10 is the most important base. We write

log x

to mean the log base ten of x.

Example:

log 10,000,000 = log 10⁷ = 7

and

log 0.00000001 = log 10^-8

Example

We can see that

log 12,343,245

is between 7 and 8 since

10,000,000 < 12,343,245 < 100,000,000

log 10,000,000 = 7

and

log 100,000,000 = 8

Example

We can see that

log 0.0000145

is between -5 and -4 since

0.00001 < 0.0000145 < 0.0001

and

log 0.00001 = -5

and

log 0.0001 = -4

Exercise

Use your calculator to find

log 1,234

and

log 0.00234

Change of base formula

We next want to be able to use our calculator to evaluate a logarithm of any base. Since our calculator can only evaluate bases e and 10, we want to be able to change the base to one of these when needed. The formula below is what we need to accomplish this task.

Change of Base Formula

\[ log_a x = \frac{log_b x}{log_b a}\]

Proof

We write

y = log_a x

So that

a^y = x

Take log_b of both sides we get

log_b a^y = log_b x

Using the power rule:

y log_b a = log_b x

Dividing by log_b a

                   log_b x
        y =
                   log_b a

Example

Find

log₂ 7

We have

                         log 7
        log₂ 7 =                = 2.807...
                         log 2

Log Equations

Example

Solve

log₂ x - log₂ (x - 2) - 3 = 0

We use the following step by step procedure:

Step 1: bring all the logs on the same side of the equation and everything else on the other side.

log₂ x - log₂(x - 2) = 3

Step 2: Use the log rules to contract to one log

                    x
        log₂              = 3
                  x - 2

Step 3: Exponentiate to cancel the log (run the hook).

               x
                        = 2³ = 8
            x - 2

Step 4: Solve for x

x = 8(x - 2) = 8x - 16

7x = 16

                 16
        x =
                  7

Step 5: Check your answer

log₂ (16/7) - log₂ (16/7 - 2) = 3

Exercises:

log(x + 2) - log(x - 1) = 1
log₂(x) + log₂(x + 5) = 2

Exponential Equations

Example

Solve for x in

2^{x - 1} = 3^{x + 1}

Step 1: Take logs of both sides using one of the given bases

log₂ 2^{x - 1} = log₂ 3^{x + 1}

Step 2: Use the log rules to simplify

x - 1 = (x + 1) log₂ 3 = (x + 1)(log 3)/(log2) = 1.58(x + 1)

Step 3: Solve for x

x - 1 = 1.58 x + 1.58

-.58 x = 2.58

x = -4.45

Step 4: Check your answer.

Exercises

3^{x - 2} = 5^2x^{- 3}
2^{1 - x} = 3^{x - 1}

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