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Exponential and Logarithmic Equations - Download as PowerPoint

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  • pg 1
									Exponential and
  Logarithmic
   Equations
Objectives:

1.Solve simple exponential and logarithmic
equations
2.Solve more complicated exponential equations
3.Solve more complicated logarithmic equations
4.Use exponential and logarithmic equations to
model and solve real-life applications
         WHY???
Applications of exponential
and logarithmic equations are
found in consumer safety
testing. For example, a
logarithmic function can be
used to model crumple zones
for automobile crash test
                           Properties
     • One-to-One Properties
         a x  a y if and only if x = y
         log a x  log a y if and only if x = y

     • Inverse Properties

         a   log a x
                       x
         log a a  x   x
  Strategies for Solving Exponential
      and Logarithmic Equations
• Rewrite the original equation in a form that allows the
  use of the One-to-One Properties of exponential or
  logarithmic functions

• Rewrite an exponential equation in logarithmic form and
  apply the Inverse Property of logarithmic functions

• Rewrite a logarithmic equation in exponential form an
  apply the Inverse Property of exponential functions
               If   au    =   av,    then u = v
This says that if we have exponential functions in equations and we can
write both sides of the equation using the same base, we know the
exponents are equal.

                        The left hand side is 2 to the something. Can we re-
     
  2 
    x
    34
    8                   write the right hand side as 2 to the something?



                     Now we use the property above. The bases are both 2
     
 2 
   2x
    34              3so the exponents must be equal.


 3
x 
34                      We did not cancel the 2’s, We just used the property
                        and equated the exponents.


You could solve this for x now.
                                   1
                               4   x         The left hand side is 4 to the
  Let’s try one more:
                                              something but the right hand

We could however re-write both the 8          side can’t be written as 4 to the
                                              something (using integer
left and right hand sides as 2 to the         exponents)
something.




           
          2 2  2x             
                              3

                                 So now that each side is written with the
                         
         2
                                 same base we know the exponents must be
            x
            2            3
          2                      equal.

                                               3
       x
       23                        Check:
                                        4    
                                               
                                               1
                                               2

             3                   1 1         1 8 1
        x                                     
                                            4 8
                                  3
             2                    2
                                      8   2    3
                                4
EXAMPLES: Solving Simple Equations
  2  64
   x


  ln x  ln 5  0
       x
  1 
     9
  3 
  e  12
   x


  ln x  4
  log10 x  2
            Steps for Solving an
           Exponential Equation



       Solve for the variable. If x is in
      more than one term get x terms on
        one side and factor out the x

        Use the 3rd Property of Logs to
        move the exponent out in front
If you have au = bv and you can’t express a and b
  with the same base, take the log of both sides
                     (ln or log)
                                                          
  25 x1
        
       2 ln
      ln 5        x
                  23                        1
                                            x                x
                                                             23


     ln
  
    ln
   12 5
      x
   x2 3
   5
    n
  x 5
  l 23
   2ll
    l2x
      nn  n

          Solve for the variable. If x is in more than one
         term get x terms on one side and factor out the
                                 x



          Use the 3rd Property of Logs to move the
                    exponent out in front

If you have au = bv and you can’t express a and b with the same
                 base, take the log of both sides
                            (ln or log)
             
   2
    5x1      x
             23

       5
       n
     x 5
      l 23
      2ll
       l2
        n
        xnn
        3
         n
      lx 
     xl 2
      2l l
       2 n
       n5 5
        n
            ln
        2 2
         2
         
        ln 
          53
           ln
        x ln 5
          
         2 ln
         2ln 5
          52
        ln 2 ln
put in calculator          3
                           2 5
                          lnln
making sure to           
                         x    
                              18
                               .
                               2
enclose the                2
                           2 5
                          lnln
numerator in
parenthesis and
                         Solve for the variable. If than one
                      Solve for the variable. If x is in morex is in
the denominator in   term get x terms on one side and factor out the
parenthesis           more than one term get x terms on
                                             x
                          one side and factor out the x
         Solving Exponential
              Equations
4  72
 x
                       3  729
                        x




3(2 )  42
     x
                       2(5 )  32
                            x
If you have an exponential equation with a base “e”
then isolate the e meaning get rid of other terms and
coefficients and then take the ln of both sides.
Remember that e’s and ln’s are inverses and “undo”
each other so they’ll cancel out and you can solve from
there.
                                             
  3 6
     x
     2
   e 2
      1
                     isolate the “e”
                                        3 
                                        e 4 x
                                            21


    2x   4
   e  1      take the ln of both sides

          3                             4
                         2x 1 ln
      2    4                           3
  ln 
    e  x1
           ln                  4
            3              ln 1
                         x         3    .644
                                    2
      Solving an Exponential
             Equation
e  5  60
 x
                        e  50
                         2x




23  2t5
             4  11   84   62x
                                      13  41
  Solving an Exponential
Equation of Quadratic Type
                     Hint: Factor
e  3e  2  0
 2x       x




e  5e  6  0
 2x       x




e 9e  36  0
 2x   x
               Steps for Solving a
              Logarithmic Equation


                              CHECK! to make sure your
                                 answer is “legal”

                    Solve for the variable. If x is in
                   more than one term get x terms on
                     one side and factor out the x

               Re-write the log equation
                 in exponential form

If the log is in more than one term,
   use log properties to condense
The secret to solving log equations is to re-write the
log equation in exponential form and then solve.


   2
   Convert
  2x 
    1
 log3 this to exponential form
                           check:

     
      3
    2 21
      x                            
                                   7
                                   3
                                      
                               log  1
                                 22
                                  2 
    
    8x
     21                            
    72x                               3
                                     2 
                                      8
                                    log
    7
      x                       This is true since 23 = 8

    2
                 4 l
                  x
                 oo g
                    g1
                l x4 3
use the first property of
logs to “condense”                        
                                          
                                        4x 
                                        x3
                                       log 1
under one log
log a 4 x
   log 
 MN 1
 a  log
    M
    aN x3                                           

                     Re-write the log equation in
                         exponential form



If the log is in more than one term, use log
           properties to condense
 4 l    x
 oo g 1  3
   
    g1 
l x4 3 4 x
    x
     3 
  4x 3 0  4
     x  xx  2                                  2


x 
    x1
 4 1 x, 
 
  x 0  4 
                        Solve for the variable. If x is in more than one
                       term get x terms on one side and factor out the
                                               x


                     Re-write the log equation in
                         exponential form



If the log is in more than one term, use log
           properties to condense
  4 l
  oo g 4 
    x1
 l x4 3 x, 
     x
     g1

                                   l 4
Remember that the
domain of logs is                    
                                     lg 1
                                    oo g
                                     44
                                       3               4
numbers greater than 0
so we need to make               not “illegal” since first log is of 4 and
sure that if we put our          second is of 1
answers back in for x
we won’t be trying to
take the log of 0 or a
                                
                              3
                              log
                              1
                               
                                1
                             log 1
                               4                     4
negative number.        ah oh---can’t take the log of - 1 or - 4 so must
                             throw this solution out




             CHECK! to make sure your
                answer is “legal”
     Solving a Logarithmic
           Equation
ln x  2


ln x  7


log 3 (5x 1)  log 3 (x  7)


log10 (x  6)  log10 (2x  1)
     Solving a Logarithmic
           Equation
5  2ln x  4     2  6ln x  10



2log 5 3x  4     log10 3z  2
      Checking for Extraneous
             Solutions
     • Make sure you check your answers because
       not all results will be solutions to the equation

           log10 5x  log10 (x 1)  2




                 Application #1
• You have deposited $500 in an account that pays 6.75%
  interest, compounded continuously. How long will it
  take to double?

   A  Pe   rt
               Application #2
• Determine the amount of time it would take $1000 to
  double in an account that pays 6.75% interest,
  compounded continuously. How does this compare to
  “application #1”?
                 Application #3
• The effective yield of a savings plan is the percent
  increase in the balance after 1 year. Find the effective
  yield for each savings plan when $1000 is deposited in a
  savings account.
  •   7% annual interest rate, compounded annually
  •   7% annual interest rate, compounded continuously
  •   7% annual interest rate compounded quarterly
  •   7.25% annual interest rate, compounded quarterly

Which savings plan has the highest effective yield? Which
 savings plan will have the highest balance after 5 years?
                Application #4
• For selected years from 1980 to 2000, the average salary
  for secondary teachers y (in thousands of dollars) for the
  year t can be modeled by the equation

                                           ( 10 ≤ t ≤ 30)
     y  38.8  23.7ln t
where t = 10 represents 1980. During which year did the
 average salary for secondary teachers reach 2.5 times its
 1980 level of $16.5 thousand?

								
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