New Topic Exponents Exponential Equations I Exponential Functions The Exponential Function Base e The Logarithm Function Logarithmic & Exponential Equations Applications:Growth & Decay Review&Test

UNIT 5  : EXPONENTIAL & LOGARITHMIC FUNCTIONS

LESSON 2: EXPONENTIAL EQUATIONS

Exponential Equations:

Definition: An exponential equation is an equation where the variable is in the exponent.

Examples of Exponential Equations:  2x = 64;    92x – 1 = 729;      500(1.02)x – 1 = 897.56

Theorem:  If  ax = ay, then x = y.    In words:  If an exponential equation has the bases equal, then the exponents must be equal.

This theorem gives us our strategy for solving exponential equations, namely convert each side of the equation to a common base.

Note:  Exponential equations may also be solved by taking logarithms of both sides of the equation.  This method will be discussed in lesson 6

Example 1:  Solve for x.

Solutions:

Example 2:  Solve for x.  Check # b.

Solutions:

APPLICATIONS:

Exponential Growth and Radioactive Decay are applications of exponential equations.

Example 1:

A bacteria culture doubles in size every 10 minutes.   It’s growth is measured by the following formula:

,  where

·        A is the number of bacteria after the given time frame

·        A0 is the starting number of bacteria

·        2 is the growth factor

·        t is the total time elapsed in the experiment

·        d is the doubling period

How many bacteria will there be in the culture after 1 ½ hours if there were 20 bacteria in the original culture?

Solution:

 A  = ? A0 = 20    t = 1 ½ h = 90 min   d = 10 min.

Therefore there will be 10 240 bacteria in the culture after 1 ½ hours.

Note:  The half-life of a radioactive substance is the period of time a given amount will decay to half of it’s original amount.

Example 2:

The half-life of radioactive radon is 4 days.  I t decays according to the formula below:

,  where

·        A is the mass remaining after the decay period

·        A0 is the original mass of radioactive material

·        ½ is the decay factor

·        t is the total time elapsed

·        h is the half-life of the material

If the amount remaining after 40 days is 6.5 g, calculate the original amount.

Solution:

 A = 6.5 g A0 = ?   t = 40 days   h = 4 days

Therefore the original mass was 6656 g.

Note: An equivalent formula for radioactive decay is: