21

LOGARITHMIC AND EXPONENTIAL FUNCTIONS

Exponential functions

Inverse relations

Exponential and logarithmic equations

Creating one logarithm from a sum

THE LOGARITHMIC FUNCTION WITH BASE b is the function

y   =  log_b 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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To cover the answer again, click "Refresh" ("Reload").

The logarithm of 1 is 0.  y = log_b1 = 0.

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

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

Proper fractions.

•  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).

Exponential functions

By definition:

log_by = x   means   b^x = y.

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

y   =  b^x.

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) = e^x.  Write the function f(−x).

f(−x)   =  e^−x

The argument x is replaced by −x.

b)   What is the relationship between the graph of  y = e^x  and the graph
b)    of  y = e^−x ?

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

c)   Sketch the graph of y = e^−x.

Inverse relations

Exponential functions and logarithmic functions with base b are inverses.

The functions  log_bx  and  b^x  are inverses.

Here are the inverse relations.  In any base b:

i)   b^log_bx = x,

and

ii)   log_bb^x = x.

Rule i) embodies the definition of a logarithm:  log_bx 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) = b^x   and   g(x) = log_bx.

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) = b^x   and   g(x) = log_bx

are inverses.

Problem 3.   Evaluate the following.

Problem 4.  

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

y = e^x.

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

f(g(x)) = ln e^x = x,

g(f(x)) = e^{ln x} = x.

Here are the graphs of  y = e^x   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.

Problem 6.   Evaluate  ln e^{arccos (−1)}.

ln e^{arccos (−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 :

5^{x + 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.

In general, if we have any equation,

Example 3.   Solve for x :

2^{x − 4}  =  3^x

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

log₂3 is some number.  The equation is solved.

Problem 7.   Solve for x :

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

10^{3x − 1}  =  2^{2x + 1}

Problem 9.   Solve for x :

Example 4.   Solve for x:

log₅(2x + 3)  =  3      

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

Problem 10.   Solve for x :

Problem 11.   Solve for x :

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:

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

Problem 12.   Solve for x:

ln (5x − 1) = ln (2x + 8).

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.

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).

Example 7.   By means of Rule i above --

n   =  log_bbⁿ,

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

For example,

Problem 15.   

Example 8.   Write the following as one logarithm:

log_bx  +  n

Problem 16.   Write as one logarithm:  log 2  +  3

Problem 17.   Write as one logarithm:  ln A  −  t

Problem 18.   Solve for x:

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	=	e − ex
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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