LOGARITHMIC AND 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?
To see the answer, pass your mouse over the colored area.
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?
• 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).
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.
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 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.
a) Let f(x) = ex. 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 = ex and the graph
y = e−x is the reflection about the y-axis of y = ex.
c) Sketch the graph of y = e−x.
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,
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.
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
Problem 3. Evaluate the following.
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
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.
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.
In general, if we have any equation,
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:
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.
103x − 1 = 22x + 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 -- 5x -- 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).
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 = logbbn,
-- we can write any number as a logarithm in any base.
Example 8. Write the following as one logarithm:
logbx + n
Problem 16. Write as one logarithm: log 2 + 3
Problem 17. Write as one logarithm: ln A − t
Problem 18. Solve for x:
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.
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.
Copyright © 2018 Lawrence Spector
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