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3.5 Solving exponential and logarithmic equations We explore some results involving exponential equations and logarithms. 3.5.1 Solving log and exponential equations In this section we concentrate on using logarithms to solve exponential equations. As a general principle, whenever we seek the value of a variable in an equation: If the variable appears as an exponent, we should think about using logarithms. Here is a set of sample problems. (The first four problems are from “Example 2” in Dr. Paul’s online math notes on logarithms at Lamar University.) Example Solve the following exponential equations for x. 1. 7x = 9 2. 24y+1 − 3y = 0. 3. et+6 = 2. 4. 5e2z+4 − 8 = 0 5. 105x−8 = 8. Solutions. In each case, since we are solving for a variable in the exponent, we may take a logarithm of both sides of the equation. In most cases, the base of the logarithm is irrelevant but in problems (3) and (4) we might as well use base e; in problem (5) we take the logarithm base 10. 1. Apply ln() to both sides of 7x = 9 to obtain ln(7x ) = ln(9) and so (by the “exponent” property of logs) x ln(7) = ln(9) and so x = ln 9 . ln 7 2. Rewrite 24y+1 − 3y = 0 as 24y+1 = 3y . Apply ln() to both sides of the equation to obtain ln(24y+1 ) = ln(3y ) and pull out the exponents (4y + 1) ln 2 = y ln 3. We need to isolate y so move part of the lefthand side over to the right: 4y ln 2 = y ln 3 − ln 2 and also the term of the righthand side involving y needs to be moved to the left: 4y ln 2 − y ln 3 = − ln 2. 132 Now factor out y: y(4 ln 2 − ln 3) = − ln 2. Solve for y by dividing both sides by the constant 4 ln 2 − ln 3 to get y= − ln 2 . 4 ln 2 − ln 3 If you want to show off to your friends (and mess up your instructor’s key!) note that 4 ln 2 − ln 3 = 16 and so you could also write down the answer ln 16 − ln 3 = ln 3 y= − ln 2 . ln( 16 3 ) (There is no real way to simplify this. It looks tempting to try to “simplify” some terms but the function ln() does not allow this.) 3. Apply ln() to both sides of et+6 = 2 to obtain t + 6 = ln 2 so t = ln 2 − 6. 4. Apply ln() to both sides of 5e2z+4 = 8 to obtain ln 5e2z+4 = ln 8 Use the “multiplication” property of logs to rewrite this as ln 5 + ln e2z+4 = ln 8 and so ln 5 + 2z + 4 = ln 8 and so 2z = ln 8 − ln 5 − 4. Therefore z = ln 8 − ln 5 − 4 . 2 5. Apply log() to both sides of 105x−8 = 8 to obtain 5x − 8 = log 8 and so x = log 8 + 8 . 5 (Note that here we are using 10 as the base of our logs.) Some more worked problems. Here are some problems off of an old exam: Solve for x in the following equations, finding the exact value of x. Then use your calculator to approximate the value of x to four decimal places. 1. 2x = 17 2. 2x = 3x+1 Solution. 1. x = log2 (17) = ln 17 ≈ 4.08746284. ln 2 133 2. Take the natural log of both sides of the equation 2x = 3x+1 to obtain ln(2x ) = ln(3x+1 ) Rewrite this as x ln 2 = (x + 1) ln 3 so x ln 2 = x ln 3 + ln 3. Move terms with an x to the leftside: x ln 2 − x ln 3 = ln 3 and factor out x. x(ln 2 − ln 3) = ln 3. Finally, solve for x: x= ln 3 ≈ −2.70951129 ln 2 − ln 3 Sometimes our equation explicitly involves logarithms and we need to use properties of logarithms to get the problem into the correct form where we can easily solve it. Here is an example: Solve the equation log3 (2x2 − 8) − log3 (x − 2) = 4. Solution. We solve log3 (2x2 − 8) − log3 (x − 2) = 4 by using our “division property” (equation 19) to 2x2 − 8 rewrite the lefthand side as log3 . We may factor 2x2 − 8 as 2(x − 2)(x + 2) and so (as long as x−2 2x2 − 8 x 6= 2) simplify = 2(x + 2) = 2x + 4. So out equation simplifies to x−2 log3 (2x + 4) = 4. We rewrite this log equation into exponential form, removing the logarithm from the problem. 2x + 4 = 34 and so 2x + 4 = 81. We can (easily!) solve nice linear equations like 2x + 4 = 81 and get x = 3.5.2 77 . 2 Finding the number of digits in an expression Modern scientific computations sometimes involve large numbers (such as the number of atoms in the galaxy or the number of seconds in the age of the universe.) Some numbers are so large it is difficult to even figure out the number of digits. Here is an example. How many decimal digits are there in the number 2300 3100 ? We can answer this question by computing the logarithm of the number base 10 and correctly interpreting the result. For example, if N is a positive integer then the number of decimal digits in 10N is N + 1 since 10N is written as a one followed by N zeroes. Since the logarithm base 10 of 10N is just N , we see that the number of decimal digits in a number is one more than the floor of the logarithm base ten. For example, log(2300 3100 ) = log(2300 ) + log(3100 ) = 300 log 2 + 100 log 3. 134 We can approximate log 2 ≈ 0.30103 and log 3 ≈ 0.47712 to write 300 log 2 + 100 log 3 ≈ 300(0.30103) + 100(0.47712) = 90.309 + 47.712 = 138.021. This tells us that 2300 3100 ≈ 10138.021 . Note that 101 = 10 has two decimal digits, 102 = 100 has three decimal digits and in general, if we want the decimal digits of an expression of the form 10x , we need to round up. So 10138.021 has 139 decimal digits. We can say more. We can approximate 10138.021 = 100.021 × 10138 ≈ 1.05 × 10138 . So 2300 3100 begins “105...” and continues with another 136 decimal places!! The largest known prime number. The set of prime numbers is an infinite set4 . However, we only know, at this time, a finite number of prime numbers. As of June 2013, the largest known prime number (according to this Wikipedia article) is 257,885,161 − 1. This is a big number! Suppose I want write out this big prime number. How many decimal digits would it take? Solution. log10 (257,885,161 ) = 57885161 · log10 (2) = 57885161( ln 2 ) ≈ 17425169.76484. ln 10 This means that 257,885,161 ≈ 1017425169.76484 = (100.76484 )(1017425169 ) ≈ 5.81887 × 1017425169 . In other words, 257,885,161 begins with a 5 and is followed by another 17425169 digits, so it has 17,425,170 digits! That’s over 17 million digits! I certainly don’t want to try to write that out! Indeed, it might be hard to get a computer system to write that out, although you could give WolframAlpha a try. In the solution to this problem on prime numbers, I calculated the number of digits in 257,885,161 . But the prime number we were after is really 257,885,161 − 1. Is it obvious that subtracting 1 from 257,885,161 won’t change the number of digits? 3.5.3 Other resources for solving exponential and logarithmic equations In the free textbook, Precalculus, by Stitz and Zeager (version 3, July 2011, available at stitz-zeager.com) this material is covered in sections 6.3 and 6.4. In the textbook by Ratti & McWaters, Precalculus, A Unit Circle Approach, 2nd ed., c. 2014 this material appears in section 3.4. In the textbook by Stewart, Precalculus, Mathematics for Calculus, 6th ed., c. 2012 (here at Amazon.com) this material appears in in section 4.5. There are a number of online resources on solving exponential and logarithmic equations. Here are some I recommend. 1. Dr. Paul’s online math notes on solving log equations, 2. Videos on solving log equations from Khan Academy. Homework. As class homework, please complete Worksheet 3.5, Solving Exponential and Log Equations available through the class webpage. 4 That there are an infinite number of primes was first proven by Euclid around 300 BC! The proof is pretty easy ... ask your instructor! 135