jdlogo

jdlogo

jdlogo

jdlogo

jdlogo

New Topic

Exponents

Exponential Equations I

Exponential Functions

The Exponential Function Base e

The Logarithm Function

Logarithmic & Exponential Equations

Applications:Growth & Decay

Review&Test

 

jdsmathnotes

 


 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.

 

Text Box: Steps for solving:
·	Convert both sides of the equation to a common base
·	Isolate the power containing the exponent
·	Equate the exponents using the above theorem
·	Solve the resulting equation
           

 

 

 

 

 

 

 

 

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:    

 

 

Return to top of page

Click here to go to homework questions