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21
LOGARITHMIC AND EXPONENTIAL FUNCTIONS
Exponential and logarithmic equations
Creating one logarithm from a sum
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.
To cover the answer again, click "Refresh" ("Reload").
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).
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.
The exponential function is the inverse of the 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) = e−x
The argument x is replaced by −x.
b) What is the relationship between the graph of y = ex and the graph
b) of y = e−x ?
y = e−x is the reflection about the y-axis of y = ex.
c) Sketch the graph of y = e−x.
Inverse relations
Exponential functions and logarithmic functions with base b are inverses.
The functions logbx and bx are inverses.
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) log 106.2
|
= 6.2 |
|
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?
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 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. 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 | |
| x − x 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 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 11. Solve for x:
ln (5x − 1) = ln (2x + 8).
| 5x − 1 | = | 2x + 8 |
| 3x | = | 9 |
| x | = | 3. |
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 12. Write as one logarithm:
k log x + m log y − n log z
Problem 13. 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 14.
| 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 15. Write as one logarithm:
log 2 + 3
| log 2 + 3 | = | log 2 + log 103 |
| = | log 2 × 103 | |
| = | log 2000. | |
Problem 16. Write as one logarithm:
ln A − t
| ln A − t | = | ln A − ln et |
| = | ln A + ln e−t | |
| = | ln Ae−t | |
Problem 17. 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 18. 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 |
| (e + 1)x | = | e |
| 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
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