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21

LOGARITHMIC AND EXPONENTIAL FUNCTIONS

Inverse relations


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

The exponential function with positive base b > 1  is the function

y  =  bx.

It is defined for every real number x.  Here is its graph:

There are two important things to note:

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

•  The 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

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

The inverse of any exponential function is a logarithmic function.  For, 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 before  (Topic 20).

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)  log 106.2  = 6.      c)  ln ex + 1  x + 1 
 
   d)  2log25  = 5       e)  10log 100  = 100       f)  eln (x − 5)  x − 5 

Problem 4.  

a)   What function is the inverse of  y = ln x (Topic 19)?

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))  =  eln x  =  x,

g(f(x))  =  ln ex  =  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 ln earccos (−1).

ln earccos (−1) = arccos (−1) = π.

See Topic 20 of Trigonometry.

Exponential and logarithmic equations

Example 2.   Solve this equation for x :

5x + 1 = 625

 Solution.   To "release" x + 1 from the exponent, take the inverse function -- the logarithm with base 5 -- of both sides.  Equivalently, write the logarithmic form (Topic 20).

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

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   =   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 6.   Solve for x :

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

Problem 7.   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 8.   Solve for x :

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

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 9.   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 10.   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 base 10, then according to the inverse relations:

2x + 1 = 11.
 
That implies
x = 5.

Problem 11.   Solve for x:

ln (5x + 1) = ln (2x − 8).

If we let each side be the exponent with base e, then

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

Skill in Algebra, Lesson 9.

One logarithm

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 12.   Write as one logarithm:

k log x  +  m log y  −  n log z


Example 7.   According to this rule,

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 13.   

   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 14.   Write as one logarithm:

log 2  +  3

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

Problem 15.   Write as one logarithm:

ln A  −  t

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

Problem 16.   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, 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 −4 is not in the domain of log2x.

Problem 17.   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:  Factorials

First Lesson on Logarithms


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