21 LOGARITHMIC AND EXPONENTIAL FUNCTIONSExponential 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. 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 functionsBy 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) = 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. 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 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.
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 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. For example,
Problem 15.
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:
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.
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 Copyright © 2018 Lawrence Spector Questions or comments? E-mail: themathpage@yandex.com |