21 LOGARITHMIC AND EXPONENTIAL FUNCTIONS Exponential and logarithmic equations 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 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) = e−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 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.
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 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).
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 6. Solve for x :
Problem 7. Solve for x. The solution may be expressed as a logarithm. 103x − 1 = 22x + 1
Problem 8. 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 9. Solve for x :
Problem 10. Solve for x :
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:
Problem 11. Solve for x: ln (5x + 1) = ln (2x − 8). If we let each side be the exponent with base e, then
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
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,
Problem 13.
Example 8. Write the following as one logarithm: logbx + n
Problem 14. Write as one logarithm: log 2 + 3
Problem 15. Write as one logarithm: ln A − t
Problem 16. Solve for x:
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.
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. Please make a donation to keep TheMathPage online. Copyright © 2001-2007 Lawrence Spector Questions or comments? E-mail: themathpage@nyc.rr.com |