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21

LOGARITHMIC AND EXPONENTIAL FUNCTIONS

Exponential functions

THE LOGARITHMIC FUNCTION WITH BASE b is the function

y  =  logb x.

b is normally a number greater than 1 (although it need only be greater than 0 and not equal to 1).  The function is defined for all x > 0.  Here is its graph for any base b.

Note the following:

•  For any base, the x-intercept is 1.  Why?

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The logarithm of 1 is 0.  y =logb1=0.

•  The graph passes through the point (b, 1).   Why?

The logarithm of the base is 1.  logbb=1.

•   The graph is below the x-axis -- the logarithm is negative -- for
 
  0 < x < 1.
 
  Which numbers are those that have negative logarithms?

Proper fractions.

 
•   The function is defined only for positive values of x.
 
  logb(−4), for example, makes no sense.  Since b is always positive, no power of b can produce a negative number.

•  The range of the function is all real numbers.

•  The negative y-axis is a vertical asymptote (Topic 18).

Example 1.   Translation of axes.   Here is the graph of the natural logarithm,  y=ln x  (Topic 20).

And here is the graph of   y=ln (x − 2) -- which is its translation 2 units to the right.

The x-intercept has moved from 1 to 3.  And the vertical asymptote has moved from 0 to 2.

Problem 1.   Sketch the graph of y=ln (x + 3).

This is a translation 3 units to the left.   The x-intercept has moved from 1 to −2.  And the vertical asymptote has moved from 0 to −3.

Exponential functions

By definition:

logby=x   means   bx=y.

Corresponding to every logarithm function with base b, we see that there is an exponential function with base b:

y  =  bx.

An exponential function is the inverse of a logarithm function. We will go into that more below.

An exponential function is defined for every real number x.  Here is its graph for any base b:

There are two important things to note:

•  The y-intercept is at (0, 1).  For, b0=1.

•  The negative x-axis is a horizontal asymptote.  For, when x is a large negative number -- e.g. b−10,000 -- then y is a very small positive number.

Problem 2.

a)   Let f(x)=ex.  Write the function f(−x).

f(−x)  =  ex

The argument x is replaced by −x.

b)   What is the relationship between the graph of  y = ex  and the graph
b)    of  y =ex ?

y = e−x is the reflection about the y-axis  of y = ex.

c)   Sketch the graph of y = ex.

Inverse relations

Exponential functions and logarithmic functions with base b are inverses.

The functions  logbx  and  bx  are inverses.

Here are the inverse relations.  In any base b:

i)   blogbx=x,

and

ii)   logbbx=x.

Rule i) embodies the definition of a logarithm:  logbx is the exponent to which b must be raised  to produce x.

Rule ii) we have seen in the previous Topic.

Now, let

f(x)=bx   and   g(x)=logbx.

Then Rule i) is  f(g(x))  =  x.

And Rule ii) is  g(f(x))  =  x.

These rules satisfy the definition of a pair of inverse functions (Topic 19).   Therefore for any base b, the functions

f(x)=bx   and   g(x)=logbx

are inverses.

Problem 3.   Evaluate the following.

   a)  log225  = 5       b)  log552x  = 2x       c)  log 106.2  = 6.
 
   d)  2log25  = 5       e)  5log5x − 1  x − 1       f)  10log 100  = 100 

Problem 4.  

a)   What function is the inverse of  y=ln x?

y=ex.

b)   Let  f(x)=ln x   and  g(x)=ex,   and show that f and g satisfy the
b)   inverse relations.

f(g(x)) = ln ex = x,

g(f(x)) = eln x = x.

Here are the graphs of  y=ex   and   y=ln x :

As with all pairs of inverse functions, their graphs are symmetrical with respect to the line  y=x.  (See Topic 19.)

Problem 5.   Evaluate the following.

   a)  ln ex + 1  x + 1       b)  eln (x − 5)  = x − 5 

Problem 6.   Evaluate  ln earccos (−1).

ln earccos (−1)=arccos (−1)=π.

"The angle whose cosine is −1 is π."

Topic 19 of Trigonometry.

Exponential and logarithmic equations

Example 2.   Solve this equation for x :

5x + 1=625.

 Solution.   When the unknown x appears as an exponent, then to "free" it, take the inverse function of both sides.

In this example, take the logarithm with base 5 of both sides.

log55x + 1  =  log5625
 
x + 1  =  log5625
 
x + 1  =  4
 
x  =  3.

In general, if we have any equation,

f(x) = a,
 
  then if g is the inverse of f:
x = g(a).

Example 3.   Solve for x :

2x − 4  =  3x

 Solution.   We may take the log of both sides either with the base 2 or the base 3.  Let us use base 2:

log22x − 4  =  log23x
 
x − 4  =  log23x
 
x − 4  =  x log23,   according to the 3rd Law
 
xx log23  =  4
 
x(1 − log23)  =  4
 
x  =         4      
1 − log23

log23 is some number.  The equation is solved.

Problem 7.   Solve for x :

2x − 5  =  32
 
log22x − 5  =  log232
 
x − 5  =  5
 
x  =  10

Problem 8.   Solve for x.  The solution may be expressed as a logarithm.

103x − 1  =  22x + 1

log 103x − 1  =  log 22x + 1
 
3x − 1  =  (2x + 1) log 2
 
3x − 1  =  2x log 2 + log 2
 
3x − 2x log 2  =  1 + log 2
 
x(3 − 2 log 2)  =  1 + log 2
 
x  =   1 + log 2 
3 − 2 log 2

Problem 9.   Solve for x :

esin x  =  1
 
ln esin x  =  ln 1
 
sin x  =  0
 
x is the radian angle whose sine is 0:
 
x  =  nπ.

Example 4.   Solve for x:

log5(2x + 3)  =  3      

Solution.   To "free" the argument of the logarithm, take the inverse function -- 5x -- of both sides.  That is, let each side be the exponent with base 5.  Equivalently, write the exponential form.

2x + 3  =  53
 
2x  =  125 − 3
 
2x  =  122
 
x  =  61.

Problem 10.   Solve for x :

log4(3x − 5)  =  0.
 
     If we let each side be the exponent with base 4, then
 
3x − 5  =  40 = 1
 
3x  =  6
 
x  =  2.

Problem 11.   Solve for x :

log2(x2 + 7)  =  4.
 
x2 + 7  =  24 = 16
 
x2  =  16 − 7 = 9
 
x  =  ±3.

Example 5.   Solve for x:

log (2x + 1) = log 11.

Solution.  If we let each side be the exponent with 10 as the base, then according to the inverse relations:

2x + 1 = 11.
  That implies
x = 5.

We may conclude, then, that if an equation looks like this:

  logb A = logb B,
then  
  A = B.

Problem 12.   Solve for x:

ln (5x − 1) = ln (2x + 8).

5x − 1  =  2x + 8
 
3x  =  9
 
x  =  3.

Skill in Algebra, Lesson 9.

Creating one logarithm from a sum

Example 6.   Use the laws of logarithms (Topic 20) to write the following as one logarithm.

log x  +  log y  −  2 log z.

  Solution.     log x  +  log y  −  2 log z  =  log  xy  −  log z2
 
   =  log  xy
z2

Problem 13.   Write as one logarithm:

k log x  +  m log y  −  n log z

Problem 14.   Write as one logarithm:

log (2x − 8) − log (x2 − 16).

log (2x − 8) − log (x2 − 16)  =  log  2x − 8
x2 − 16
 
   =  log       2(x − 4)    
(x − 4)(x + 4)
 
   =  log      2   
x + 4

Example 7.   By means of Rule i above --

n  =  logbbn,

-- we can write any number as a logarithm in any base.

For example,

7  =  log227
 
5.9  =  log335.9
 
t  =  ln et
 
3  =  log 1000

Problem 15.   

   a)   2 = ln e2   b)   1 = ln e

Example 8.   Write the following as one logarithm:

logbx  +  n

  Solution. logbx  +  n  =  logbx  +  logbbn
 
   =  logbxbn.

Problem 16.   Write as one logarithm:

log 2  +  3

log 2  +  3  =  log 2  +  log 103
 
   =  log 2 × 103
 
   =  log 2000.

Problem 17.   Write as one logarithm:

ln A  −  t

ln A  −  t  =  ln A  −  ln et
 
   =  ln A  +  ln et
 
   =  ln Aet

Problem 18.   Solve for x:

log2x  +  log2(x + 2)  =  3.
 
log2[x(x + 2)]  =  3.
 
    If we now let each side be the exponent with base 2
    (or, write 3 = log223), then
 
x(x + 2)  =  23 = 8.
 
x2 + 2x − 8  =  0
 
(x − 2)(x + 4)  =  0
 
x  =  2 or −4.

See Skill in Algebra, Lesson 37.

We must reject the solution x = − 4, however, because the negative number −4 is not in the domain of log2x.

Problem 19.   Solve for x.

ln (1 + x) − ln (1 − x)  =  1.
 
 =  1.
 
    If we now let each side be the exponent with base e, then
 
 =  e
 
1 + x  =  eex
 
ex + x  =  e − 1
 
(e + 1)x  =  e − 1
 
x  = 

The student can now begin to see:  To solve any equation for the argument of a function, take the inverse function of both sides.

This Topic concludes our study of functions and their graphs.


Next Topic:  Sigma notation for sums

First Lesson on Logarithms


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