LOGARITHMIC AND EXPONENTIAL FUNCTIONS Inverse relations Exponential functions Exponential and logarithmic equations One logarithm 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? | 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 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 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 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. 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 ? ) | = x ? 5 | 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 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) = ?. 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). log55>x + 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(x? + 7)| = | 4| | x? + 7| = | 24 = 16| | x? | = | 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. | Problem 11. Solve for x: ln (5x ? 1) = ln (2x + 8). If we let each side be the exponent with base e, then 5x ? 1| = | 2x + 8| | 3x| = | 9| | x| = | 3. | Skill in Algebra, Lesson 9. 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 Solution. log x + log y ? 2 log z| = | log | xy ? log z? | | | = | log | xy z? | Problem 12. Write as one logarithm: k log x + m log y ? log z Example 7. According to this rule, 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 13. a) | 2 = ln e? | | b) | 1 = ln e| Example 8. Write the following as one logarithm: logbx + n Solution. | logbx + n| = | logbx + logbbn | | | = | logbxbn | Problem 14. Write as one logarithm: log 2 + 3 log 2 + 3| = | log 2 + log 103 | | | = | log 2 ? 103| | | = | log 2000 | Problem 15. Write as one logarithm: ln A ? t ln A ? t| = | ln A ? ln et | | = | ln A + ln e? t | | | = | ln Ae? t| Problem 16. 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, then | | x(x + 2) | = | 23 = 8. | | x? + 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 17. 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| = | | 20 LOGARITHMS Definition Common logarithms Natural logarithms The three laws of logarithms Proof of the laws of logarithms Change of base WHEN WE ARE GIVEN the base 2, for example, and exponent 3, then we can evaluate 23. 23 = 8. Inversely, if we are given the base 2 and its power 8 — 2? = 8 — then what is the exponent that will produce 8? That exponent is called a logarithm. We call the exponent 3 the logarithm of 8 with base 2. We write 3 = log28. We write the base 2 as a subscript. is the exponent to which 2 must be raised to produce 8. A logarithm is an exponent. Since 104 = 10,000 then log1010,000 = 4. “The logarithm of 10,000 with base 10 is 4. ” 4 is the exponent to which 10 must be raised to produce 10,000. “104 = 10,000” is called the exponential form. “log1010,000 = 4” is called the logarithmic form. Here is the definition: logbx = n means bn = x. That base with that exponent produces x. Example 1. Write in exponential form: log232 = 5 Answer. 25 = 32 Example 2. Write in logarithmic form: 4? 2 = | 1 16| . | Answer. log4| 1 16| = ? 2. | Problem 1.

Which numbers have negative logarithms? To see the answer, pass your mouse over the colored area. To cover the answer again, click “Refresh” (“Reload”). Do the problem yourself first! Proper fractions. Example 3. Evaluate log81. Answer. 8 to what exponent produces 1? 80 = 1. log81 = 0. We can observe that, in any base, the logarithm of 1 is 0. logb1 = 0 Example 4. Evaluate log55. Answer. 5 with what exponent will produce 5? 51 = 5. Therefore, log55 = 1. In any base, the logarithm of the base itself is 1. logbb = 1 Example 5. log22m = ? Answer. 2 raised to what exponent will produce 2m ? m, obviously. log22m = m.

The following is an important formal rule, valid for any base b: logbbx = x This rule embodies the very meaning of a logarithm. x — on the right — is the exponent to which the base b must be raised to produce bx. The rule also shows that the inverse of the function logbx is the exponential function bx. We will see this in the following Topic. Example 6 . Evaluate log3 | 1 9| . | Answer. | 1 9| is equal to 3 with what exponent? | 1 9| = 3? 2. | log3 | 1 9| = | log33? 2 = ? 2. | Compare the previous rule. Example 7. log2 . 25 = ? Answer. .25 = ? = 2? 2. Therefore, log2 . 25 = log22? 2 = ? 2. Example 8. og3 = ? Answer. = 31/5. (Definition of a rational exponent. ) Therefore, log3 = log331/5 = 1/5. Problem 2. Write each of the following in logarithmic form. To see the answer, pass your mouse over the colored area. To cover the answer again, click “Refresh” (“Reload”). a) bn = x | logbx = n| | b) 23 = 8 | log28 = 3| | c) 102 = 100 | log10100 = 2| | d) 5? 2 = 1/25. | log51/25 = ? 2. | Problem 3. Write each of the following in exponential form. a) logbx = n | bn = x| | b) log232 = 5 | 25 = 32| | c) 2 = log864 | 82 = 64| | d) log61/36 = ? 2 | 6? 2 = 1/36 | Problem 4.

Evaluate the following. a) log216 | = 4| | b) log416 | = 2| | c) log5125 | = 3| | d) log81 | = 0| | e) log88 | = 1| | f) log101 | = 0| Problem 5. What number is n? a) log10n = 3 | 1000| | b) 5 = log2n | 32| | c) log2n = 0 | 1 | | d) 1 = log10n | 10 | e) logn | 1 16| = ? 2 | 4 | | f) logn> | 1 5| = ? 1 | 5 | | g) log2 | 1 32| = n | ? 5 | | h) log2| 1 2| = n | ? 1 | Problem 6. logbbx = x Problem 7. Evaluate the following. a) log9| 1 9| | = log99? 1 = ? 1| b) log9| 1 81| | = ? 2| | c) log2| 1 4| | = ? 2| | d) log2| 1 8| | = ? | | e) log2| 1 16| | = ? 4| f) log10. 01 | ? 2| | g) log10. 001 | ? 3| | h) log6 | = 1/3| | i) logb | = 3/4| Common logarithms The system of common logarithms has 10 as its base. When the base is not indicated, log 100 = 2 then the system of common logarithms — base 10 — is implied. Here are the powers of 10 and their logarithms: Powers of 10: | | 1 1000| | 1 100| | 1 10| | 1| | 10| | 100| | 1000| | 10,000| | | Logarithms: | | ? 3| | ? 2| | ? 1| | 0| | 1| | 2| | 3| | 4| Logarithms replace a geometric series with an arithmetic series. Problem 7. log 10n = ? . The base is 10. Problem 8. log 58 = 1. 7634. Therefore, 101. 7634 = ? 58. 1. 7634 is the common logarithm of 58. When 10 is raised to that exponent, 58 is produced. Problem 9. log (log x) = 1. What number is x? log a = 1, implies a = 10. (See above. ) Therefore, log (log x) = 1 implies log x = 10. Since 10 is the base, x = 1010 = 10,000,000,000 Natural logarithms The system of natural logarithms has the number called “e” as its base. (e is named after the 18th century Swiss mathematician, Leonhard Euler. ) e is the base used in calculus. It is called the “natural” base because of certain technical considerations. x has the simplest derivative. See Lesson 14 of An Approach to Calculus. ) e can be calculated from the following series involving factorials: e| = 1 + | 1 1! | + | 1 2! | + | 1 3! | + | 1 4! | + . . . | e is an irrational number, whose decimal value is approximately 2. 71828182845904. To indicate the natural logarithm of a number, we use the notation “ln. ” ln x means logex. Problem 10. What number is ln e ? ln e = 1. The logarithm of the base itself is always 1. e is the base. Problem 11. Write in exponential form (Example 1): y = ln x. ey = x. e is the base. The three laws of logarithms . logbxy = logbx + logby “The logarithm of a product is equal to the sum of the logarithms of each factor. ” 2. logb| x y| = logbx ? logby| “The logarithm of a quotient is equal to the logarithm of the numerator minus the logarithm of the denominator. ” 3. logb xn = n logbx “The logarithm of a power of x is equal to the exponent of that power times the logarithm of x. ” We will prove these laws below. Example 9. Apply the laws of logarithms to log | abc? d 3| . | Answer. According to the first two laws, log | abc? d 3| = | log (abc? ) ? log d 3| | | | = | log a + log b + log c? ? log d 3| | | = | log a + log b + 2 log c ? 3 log d,| according to the third law. The Answer above shows the complete theoretical steps. In practice, however, it is not necessary to write the line log | abc? d 3| = | log (abc? ) ? log d 3| . | The student should be able to go immediately to the next line — log | abc? d 3| = | | log a + log b + log c? ? log d 3| — if not to the very last line log | abc? d 3| = | log a + log b + 2 log c ? 3 log d. | Example 10. Use the laws of logarithms to rewrite log | z5| . | Answer. | | log | z5| = log x + log ? log z5| Now, = y?. (Lesson 29 of Algebra. Therefore, according to the third law, log | z5| = log x + ? log y ? 5 log z. | Example 11. Use the laws of logarithms to rewrite ln . Solution. ln | = | ln (sin x ln x)? | | | = | ? ln (sin x ln x), 3rd Law | | | = | ? (ln sin x + ln ln x), 1st Law | Note that the factors sin x ln x are the arguments of the logarithm function. Example 12. Solve this equation for x: log 32x + 5| = | 1| | Solution. According to the 3rd Law, we may write| | (2x + 5)log 3| = | 1| | Now, log 3 is simply a number. Therefore, on distributing log 3,| | 2x· log 3 + 5 log 3| = | 1| | x· log 3| = | 1 ? 5 log 3| | x| = | 1 ? 5 log 3 2 log 3| By this technique, we can solve equations in which the unknown appears in the exponent. Problem 12. Use the laws of logarithms to rewrite the following. a) log | ab c| | = log a + log b ? log c| | b) log | ab? c4| | = log a + 2 log b ? 4 log c| | c) log | z| | = 1/3 log x + 1/2 log y ? log z| d) ln (sin? x ln x) = ln sin? x + ln ln x = 2 ln sin x + ln ln x e) ln | = ? ln (cos x· x1/3 ln x)| | | = ? (ln cos x + 1/3 ln x + ln ln x)| f) ln (a2x ? 1 b5x + 1 ) | = ln a2x ? 1 + ln b5x + 1| | | = (2x ? ) ln a + (5x + 1) ln b| Problem 13. Solve for x. ln 23x + 1| = | 5| | (3x + 1) ln 2| = | 5| | 3x ln 2 + ln 2| = | 5| | 3x ln 2| = | 5 ? ln 2| | x| = | 5 ? ln 2 3 ln 2| Problem 14. Prove: ? ln x| = ln | 1 x| . | ?ln x = (? 1)ln x| = | ln | x? 1,| Third law| | | = | ln | 1 x| | Proof of the laws of logarithms The laws of logarithms will be valid for any base. We will prove them for base e, that is, for y = ln x. 1. ln ab = ln a + ln b. The function y = ln x is defined for all positive real numbers x. Therefore there are real numbers p and q such that p = ln a and q = ln b.

This implies a = e p and b = e q. Therefore, according to the rules of exponents, ab = e p· e q = ep + q. And therefore ln ab = ln ep + q = p + q = ln a + ln b. That is what we wanted to prove. In a similar manner we can prove the 2nd law. Here is the 3rd: 3. ln an = n ln a. There is a real number p such that p = ln a; that is, a = e p. And the rules of exponents are valid for all rational numbers n (Lesson 29 of Algebra; an irrational number is the limit of a sequence of rational numbers). Therefore, an = e pn. This implies ln an = ln e pn = pn = np = n ln a. That is what we wanted to prove. Change of base

Say that we know the values of logarithms of base 10, but not, for example, in base 2. Then we can convert a logarithm in base 10 to one in base 2 — or any other base — by realizing that the values will be proportional. Each value in base 2 will differ from the value in base 10 by the same constant k. Now, to find that constant, we know that Therefore, on putting x = 2: which implies Therefore, By knowing the values in base 10, we can in this way calculate the values in base 2. In general, if we know the values in base a, then we can change to base b as follows: Problem 15. Write the rule for changing from base e to base 8.