13.3 Using Logarithms and Exponentiation to Solve Equations 451 log b x = log k x log k b Returning to our original problem, we convert log 3 100 5 to log base 10 as follows. log 3 100 5 = 1 5 log 100 log 3 = 1 5 2 log 3 = 2 5 log 3 ◆ EXAMPLE 13.5 Solve for x if log 7 (4x) = 3. SOLUTION In order to solve for x we need to free the x from the logarithm. There are two approaches we can take to this. The first is to undo log 7 by exponentiating both sides of the equation with base 7. log 7 (4x) = 3 7 log 7 (4x) = 7 3 4x = 7 3 x = 7 3 4 = 343 4 Notice that if we use a base other than 7 to exponentiate, then we will get stuck. For instance, we could write 10 log 7 (4x) = 10 3 , but we cannot simplify the expression 10 log 7 (4x) , so this course of action is unproductive. An alternative mindset for solving the original problem is to think about what log 7 (4x) = 3 means and write the statement in exponential form. log 7 (4x) = 3 means “3 is the number we must raise 7 to in order to get 4x.” So 7 3 = 4x. Then x = 7 3 4 = 343 4 . ◆ Theme and Variation. Examples 13.4 and 13.5 are prototypical examples. Example 13.4 is of the form B f(x) = A, and Example 13.5 is of the form log b f(x)=A. Knowing our way around equations of these forms will serve as a guideline for strategizing when we have more complicated examples. Below we show variations on these basic themes. Along the way, we’ll try to point out some pitfalls so you can walk around them instead of falling into them. It’s surprisingly easy either to get caught in a frenzy of unproductive manipulation or to plunge down a short, dead-end street when approaching exponential or logarithmic equations. Take time to strategize. ◆ EXAMPLE 13.6 Solve for x if 5 2x+1 = 20. SOLUTION We need to “bring down the exponent” to solve for x. Log property (iii) will help us do this. Again, although we can use log b with any base b, if we want a numerical approximation then using common logarithms or natural logarithms simplifies the task. 452 CHAPTER 13 Logarithmic Functions 5 2x+1 = 20 Take log of both sides. 8 log 5 2x+1 = log 20 Use log rule (iii); don’t forget parentheses around 2x + 1. (2x + 1) log 5 = log 20 2x log 5 + log 5 = log 20 This is just a linear equation in x. (This might not be immediately clear at first, but log 5 is just a constant, as is log 20.) Get all terms with x on one side and all else on the other. 2x log 5 = log 20 − log 5 Divide by the coefficient of x. x = log 20 − log 5 2 log 5 At this point, you could leave the answer as is, simplify, or look for a numerical approxi- mation. Even if you plan to do the latter, it still is helpful to simplify first. log 20 − log 5 2 log 5 = log(20/5) log 25 = log 4 log 25 . 9 ◆ ◆ EXAMPLE 13.7 Solve for x if 5 2x+1 − 18 = 2. CAUTION We need a strategy here. Our goal is to “bring down the 2x + 1,” but taking logs right away doesn’t help us. Although log(5 2x+1 − 18) = log 2, we can’t do much with the expression on the left because there is no log rule to simplify log(A + B). The x is trapped inside the log with no escape route. However, if we put the equation into the form B f(x) =A by adding 18 to both sides, then we’re in business. In fact, we’re back to 5 2x+1 = 20, as in Example 13.6. ◆ REMARK If A = C + D, then ln(A) = ln(C + D),but ln A = ln C + ln D. You must take the log of each entire side of the equation. As a concrete example, 2 = 1 + 1, but ln 2 = ln 1 + ln 1, ln 2 = 0. 8 An alternative strategy is to begin by dividing both sides of the equation by 5 to get 5 2x = 4. Now take logs to get log(5 2x ) = log 4, or 2x log 5 = log 4. Consequently, x = log 4 2 log 5 . 9 Alternatively, log 20 5 2 log 5 = log 4 2 log 5 = log 2 2 2 log 5 = 2 log 2 2 log 5 = log 2 log 5 . 13.3 Using Logarithms and Exponentiation to Solve Equations 453 Similarly, if A = C + D, then e A = e C+D = e C · e D ,but e A =e C +e D . As a concrete example, 2 = 1 + 1, 10 2 = 10 (1+1) = 100, but 10 2 = 10 1 + 10 1 = 20. ◆ EXAMPLE 13.8 Solve for x if 5 x+1 − 20 5 x = 0. SOLUTION Let’s strategize. We’re trying to solve for a variable in the exponent, so perhaps we can get this into the form B f(x) =A. First we’ll clean it up a bit; we can clear the denominator by multiplying by 5 x . 5 x  5 x+1 − 20 5 x  = 5 x (0) 5 2x+1 − 20 = 0 5 2x+1 = 20, and we’re back to our familiar problem. ◆ ◆ EXAMPLE 13.9 Solve for x if log 7 (x 3 ) + log 7 4 = 2 log 7 (x) + 3. SOLUTION We have plenty of options here. Below we’ll work through a couple of approaches. Approach 1. Try to put this equation into the form log b f(x)=A. log 7 (x 3 ) + log 7 4 − 2 log 7 (x) = 3 Consolidate: log 7 (x 3 ) + log 7 4 − 2 log 7 (x) = log 7 (x 3 ) + log 7 4− log 7 (x 2 ) = log 7  4x 3 x 2  log 7  4x 3 x 2  = 3 But 4x 3 x 2 = 4x; x = 0 is not in the domain of log 7 x. log 7 (4x) = 3 Now we’re back to Example 13.5. 454 CHAPTER 13 Logarithmic Functions Alternatively, we could have grouped terms as follows. 3 log 7 (x) − 2 log 7 (x) =−log 7 4 + 3 log 7 (x) =−log 7 4 + 3 Exponentiating now gives: 7 log 7 x = 7 − log 7 4+3 Careful with the right-hand side! 7 A+B = 7 A 7 B , not 7 A + 7 B . x = 7 − log 7 4 · 7 3 x = 7 log 7 (4 −1 ) · 7 3 x = (4 −1 )7 3 = 343 4 Approach 2. Exponentiate immediately. This works, but not as neatly as Approach 1. 7 [log 7 (x 3 )+log 7 4] = 7 [2 log 7 x+3] Now use the laws of logs and exponents to simplify. This is done in Example 13.3 4x 3 = x 2 7 3 We can divide through by x 2 provided x = 0. 4x = 7 3 or x = 0 x = 0 because 0 is not in the domain of the log function. x = 7 3 4 Discard the extraneous root x = 0. ◆ Two Basic Principles 1. If you want to solve for a variable and it’s caught up in an exponent, you need to bring down the variable. This requires taking the logarithm of both sides of the equation. If you can get the equation in the form B f(x) =A, you are in great shape. Taking logs of both sides leaves you with f(x)log B = log A f(x)= log A log B . Since log A log B is independent of x, it is just a constant. Don’t be fazed by its bulk; treat it as you would any other constant. You now have f(x)=k. CB f(x) =A g(x) is a nice form as well. Taking logs of both sides leaves you with log(C · B f(x) )=log(A g(x) ) log C + f(x)log B = g(x) log A. Again, log A, log B, and log C are just constants, so the equation is analogous to 3 + 5f(x)=7g(x). 2. If you want to solve for a variable and it is caught inside a logarithm, (i.e., in the argument, or input, of the log function), then you’ll need to exponentiate to undo the log. Use the same base for exponentiation as is in the log. If you can get your equation in the form log b f(x)=C, 13.3 Using Logarithms and Exponentiation to Solve Equations 455 then you can exponentiate both sides of the equation using b as a base to get b log b f(x) =b C f(x)=b C . Since b C is just a constant, you’re in good shape. The Overarching Principle for Solving Equations of the Form g(x) = Constant Suppose you have an equation of the form g(x) = k, where k is an expression without x’s. If g is invertible, you can solve for x by undoing g. g −1 ( g(x) ) = g −1 (an expression without x’s). Logarithms are undone through exponentiation, and exponentials are undone by taking logarithms, because logarithmic and exponential functions are inverses. 10 ◆ EXAMPLE 13.10 Consider the equation 1 [ln(x+2)] 3 + 7 = 34. Solve for x. SOLUTION Think of what was done to x and undo it. Remember that because you put on your socks and then your shoes, your shoes must come off before your socks. To construct the left- hand side of the equation, we begin with x and do the following: Add 2, take the natural logarithm, cube the result, take the reciprocal, add 7. To undo this sequence, we can subtract 7, take the reciprocal, take the cube root, exponentiate, subtract 2. 1 [ln(x + 2)] 3 + 7 = 34 Subtract 7 from both sides. 1 [ln(x + 2)] 3 = 27 Take the reciprocal of both sides. (If A = B, A and B = 0, then 1/A = 1/B.) [ln(x + 2)] 3 = 1 27 Take the cube root of both sides. ln(x + 2) = 1 3 Exponentiate with base e (since ln is log e ). e ln(x+2) = e 1/3 x + 2 = e 1/3 Subtract 2. x = e 1/3 − 2 ◆ Worked Examples in Solving for x (Try these on your own and then read the solutions. Notice that different strategies are adopted depending on the problem.) 10 You may be wondering why when you solve the equation B f(x) = A you can use logs with any base to bring down f(x) while to undo log b you must exponentiate with base b. The reason is that a x = b kx for the appropriate k, where b is any positive number, so a logarithm with any base will do. For example, 5 x = (10 log 5 ) x = 10 (log 5)x = 10 kx , where k is the constant log 5. Because 5 x = 10 kx , log base 10 will help you out. 456 CHAPTER 13 Logarithmic Functions (a) 3 2x+1 = 8 x Bring those exponents down; take ln of both sides. ln(3 2x+1 ) = ln 8 x (2x + 1) ln 3 = x ln 8 Multiply out to free the x. 2x ln 3 + ln 3 =x ln 8 This equation is linear in x. 2x ln 3 − x ln 8 =−ln 3 x(2ln3−ln 8) =−ln 3 x = − ln 3 2ln3−ln 8 or x = − ln 3 ln(9/8) (b) [log(x 2 + 1)] 2 = 4 Unpeel the problem as was done above in Example 13.10. log(x 2 + 1) =± √ 4 log(x 2 + 1) = 2 or log(x 2 + 1) =−2 10 log(x 2 +1) = 10 2 10 log(x 2 +1) = 10 −2 (x 2 + 1) = 10 2 (x 2 + 1) = 10 −2 x 2 = 100 −1 x 2 = 0.01 − 1 =−0.99 x =± √ 99 x 2 can’t be negative. (c) ln √ x = ln x 5 − 7 Rewrite this. (1/2) ln x = 5lnx−7 This equation is linear in ln x. (1/2) ln x − 5lnx=−7 Solve for ln x and then exponentiate. −(9/2) ln x =−7 ln x = 14/9 x = e 14/9 (d) e 2x + 3e x = 10 This equation is quadratic in e x . (e x ) 2 + 3e x − 10 =0 If you like, let u = e x and solve for u. Then return to e x and solve for x. u 2 + 3u − 10 =0 (u − 2)(u + 5) = 0 u = 2oru=−5 But u = e x . e x = 2ore x =−5 x=ln 2 or e x =−5 But e x can never be negative, so e x =−5. x = ln 2 13.3 Using Logarithms and Exponentiation to Solve Equations 457 (e) 5 ln x = 7x Bring down the exponent; take ln of both sides. ln(5 ln x ) = ln(7x) Separate out the ln x’s. ln x(ln 5) = ln 7 + ln x This equation is linear in ln x. Solve for ln x and then exponentiate. (ln 5) ln x − ln x = ln 7 ln x(ln 5 − 1) = ln 7 ln x = ln 7 ln 5−1 x = e  ln 7 ln 5−1  or x =  e ln 7  1/(ln 5−1) = 7 1/(ln 5−1) 458 CHAPTER 13 Logarithmic Functions Exploratory Problem for Chapter 13 Pollution Study 1. An environmental policy advisory board is studying information supplied to them concerning pollution levels in one of India’s rivers. The data have been presented as follows. 5 4 3 2 1 1 2 3 4567 t (in years) (7, 5) log C C = number of coliform organisms per 100 ml of water On the vertical axis is the log of the number of coliform organisms per 100 ml of water and on the horizontal axis is time, measured in years from the date the study of the river began. The data look more or less like a line passing through the points (0, 3) and (7, 5). This kind of presentation, plotting the logarithm of the pol- lution level against time, is a semilog plot. Find a formula for C(t), the number of coliform organisms in 100 ml of water, as a function of time. 2. (a) Show that in general, if a quantity grows exponentially, then the semilog plot will be a straight line. In particular, show that if a quantity grows according to the equation y = Ce kt , then ln y is a linear function of t. What is the significance of the slope of this line? The vertical intercept of the line? (b) Show that if ln y is a linear function of t,lny=mt + b, then y is an exponential function of t. Optional. Real-world data about pollution levels or population size won’t follow a formula exactly. Nevertheless, we try to model their behavior by the most appropriate function. Go find some data about a quantity you think grows (or decays) exponentially, at least over a certain time period. Plot the data points on a graph using a semilog plot. Does it seem reasonable to fit a straight line to these points? If not, then the phenomenon must not actually be exponential. If so, find the line of best fit. If you know how to get this by doing linear regression with the help of a computer or calculator, do so. Otherwise, do it by eye. Then come up with an exponential function modeling the phenomenon you have chosen. Exploratory Problem for Chapter 13 459 PROBLEMS FOR SECTION 13.3 1. Solve for t: P = P 0 e kt . 2. Solve for x: (Don’t expect “pretty” answers.) (a) 10 2x = 93 (b) 10 3x+2 = 1,000,000 (c) 2 x+1 = 7 (d) 3 x 3 x 2 = 3 (e) 5B x = (2C) x+1 (f) ln x + 2 = 5 (g) log 10 x = 17 (h) ln(5x − 40) = 3 (i) log 10 (2x 2 + 4) = 2 (j) 3 · 2 x/7 − 4 = 12 3. (a) Approximate log 3 16 (with error less than 0.005) using your calculator. (b) Rewrite log 3 16 in terms of log base 10. (c) Rewrite log 3 16 in terms of log base e. (d) Rewrite log 3 16 in terms of log base 7. 4. Solve for x. (a) 3 ln x + 5 = (ln x) ln 2 (b) 2  7 1+log x  = 8 (c) Ke x + K = Le x − L, where K and L are constants and 0 <K<L (d) R(1 + n) nx = (P n) x , where P , R, and n are constants (e) 3b x = c x 3 2x , where b and c are constants 5. Solve for x. (a) 2 x 2 2 x = 3 x (b) 3 x 2 +2x = 1 (c) 3 ln(x 4 ) − 2ln2x=10 (d) e 2x + e x − 6 = 0 (e) e x + 8e −x = 6 (f) (ln x)(ln 5) = ln 4x 6. The Richter scale, introduced in the mid-1900s, measures the intensity of earthquakes. A measurement on the Richter scale is given by M = log I S , where I is the intensity of the quake and S is some standard. Suppose we want to compare the intensity, I 1 , of a particular earthquake with the intensity, I 2 , of a less violent quake. The difference in their measurements on the Richter scale is log I 1 S − log I 2 S = log  I 1 S I 2 S  = log I 1 I 2 . In particular, suppose that one earthquake measures 7 on the Richter scale and another measures 4. Then log I 1 I 2 = 7 − 4 = 3. 460 CHAPTER 13 Logarithmic Functions Therefore, I 1 I 2 = 10 3 = 1000. The former earthquake has 1000 times the intensity of the latter. (a) On August 20, 1999, there was an earthquake in Costa Rica (50 miles south of San Jose) measuring 6.7 on the Richter scale and another in Montana (near the Idaho border) measuring 5 on the Richter scale. How many times more intense was the Costa Rican earthquake? (b) The 1989 earthquake in San Francisco measured 7.1 on the Richter scale. How many times more intense was the earthquake in Turkey on August 17, 1999, measuring 7.4 on the Richter scale? In Problems 7 through 32, solve for x. 7. 5 3x+2 = 2 5x 8. 3 2 x−3 = 7 2x+1 9. π · 3 1+2x = √ π5 x 10. e 3x = ( 5 e ) x+1 11. e 2 e x = π 3x+3 12. 3 x · 5 3 x+1 = 0 13. 7+π3 x+2 2 = 3π 14. log x − log(x + 1) = 2 15. ln x 2 = 3 + ln x 16. ln √ x + ln x 2 = 1 − 2lnx 17. [ln(2x + 3)] 2 − 9 = 0 18. log x[log(x + 3) − 2] = 0 19. e x (e x − 5) = 0 20. e x (e x − 5) = 6 21. e 2x − 4e x + 3 = 0 22. 2e 2x + 6 = 11e x 23. e −2x − e −x = 6 24. e −2x = 2 25. e x − 1 = e −x . x. SOLUTION Think of what was done to x and undo it. Remember that because you put on your socks and then your shoes, your shoes must come off before your socks. To construct the left- hand side of the. Ke x + K = Le x − L, where K and L are constants and 0 <K<L (d) R(1 + n) nx = (P n) x , where P , R, and n are constants (e) 3b x = c x 3 2x , where b and c are constants 5. Solve for x. (a). raise 7 to in order to get 4x.” So 7 3 = 4x. Then x = 7 3 4 = 343 4 . ◆ Theme and Variation. Examples 13.4 and 13.5 are prototypical examples. Example 13.4 is of the form B f(x) = A, and Example