Solving exponential and logarithmic equations We explore some results involving exponential equations and logarithms. Elementary Functions Part 3, Exponential Functions & Logarithms In this presentation we concentrate on using logarithms to solve exponential equations. Lecture 3.5a, Solving Equations With Logarithms As a general principle, whenever we seek the value of a variable in an equation: Dr. Ken W. Smith Sam Houston State University If the variable appears as an exponent, we should think about using logarithms 2013 Smith (SHSU) Elementary Functions 2013 1 / 16 Smith (SHSU) Elementary Functions 2013 2 / 16 Solving exponential and logarithmic equations Solving exponential and logarithmic equations Here is a set of sample problems. (The first four problems are from “Example 2” in Dr. Paul’s online math Example Solve the following exponential equations for x . notes on logarithms at Lamar University.) 1 7 x = 9 Example Solve the following exponential equations for x . 1 7 x = 9 Solutions. 1 Apply ln() to both sides of 7 x = 9 to obtain 2 2 4 y +1 − 3 y = 0 . 3 e t +6 = 2 . ln(7 x ) = ln(9) 4 5 e 2 z +4 − 8 = 0 5 10 5 x − 8 = 8 . and so (by the “exponent” property of logs) x ln(7) = ln(9) 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 and so x = ln 9 ln 7 . 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. Smith (SHSU) Elementary Functions 2013 3 / 16 Smith (SHSU) Elementary Functions 2013 4 / 16
Solving exponential and logarithmic equations Solving exponential and logarithmic equations Example Solve the following exponential equations for x . Example Solve the following exponential equations for x . 2 2 4 y +1 − 3 y = 0 . 3 e t +6 = 2 . 4 5 e 2 z +4 − 8 = 0 Solutions. Solutions. 2 Rewrite 2 4 y +1 − 3 y = 0 as 2 4 y +1 = 3 y . Apply ln() to both sides of 3 Apply ln() to both sides of e t +6 = 2 to obtain t + 6 = ln 2 so the equation to obtain t = ln 2 − 6 . ln(2 4 y +1 ) = ln(3 y ) 4 Apply ln() to both sides of 5 e 2 z +4 = 8 to obtain ln 5 e 2 z +4 = ln 8 and pull out the exponents (4 y + 1) ln 2 = y ln 3 . Use the “multiplication” property of logs to rewrite this as ln 5 + ln e 2 z +4 = ln 8 Isolate y . 4 y ln 2 − y ln 3 = − ln 2 . and so Factor out y : ln 5 + 2 z + 4 = ln 8 y (4 ln 2 − ln 3) = − ln 2 . and so Solve for y by dividing both sides by the constant 4 ln 2 − ln 3 to get 2 z = ln 8 − ln 5 − 4 . ln 2 ln 2 y = − 4 ln 2 − ln 3 = ln 3 − ln 16 . Smith (SHSU) Elementary Functions 2013 5 / 16 Smith (SHSU) Elementary Functions 2013 6 / 16 z = ln 8 − ln 5 − 4 . 2 Solving exponential and logarithmic equations Solving exponential and logarithmic equations 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 Example Solve the following exponential equations for x . approximate the value of x to four decimal places. 5 10 5 x − 8 = 8 . 1 2 x = 17 2 2 x = 3 x +1 Solutions. Solution. 5 Apply log() to both sides of 10 5 x − 8 = 8 to obtain 5 x − 8 = log 8 and 1 x = log 2 (17) = ln 17 ln 2 ≈ 4 . 08746284 . 2 Take the natural log of both sides of the equation 2 x = 3 x +1 to obtain so x = log 8+8 . ln(2 x ) = ln(3 x +1 ) 5 (Note that here we are using 10 as the base of our logarithm in this Use the exponent property to pull down the exponents and then problem.) isolate x : x ln 2 = ( x + 1) ln 3 x ln 2 = x ln 3 + ln 3 . x ln 2 − x ln 3 = ln 3 x (ln 2 − ln 3) = ln 3 . Smith (SHSU) Elementary Functions 2013 7 / 16 Smith (SHSU) Elementary Functions 2013 8 / 16 ln 3
Solving exponential and logarithmic equations Exponential Functions 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 In the next presentation we continue to practice applications of logarithms. log 3 (2 x 2 − 8) − log 3 ( x − 2) = 4 . (END) Solution. We solve log 3 (2 x 2 − 8) − log 3 ( x − 2) = 4 by using our quotient 2 x 2 − 8 x − 2 . We may factor 2 x 2 − 8 property to rewrite the lefthand side as log 3 as 2( x − 2)( x + 2) and so (as long as x � = 2 ) simplify 2 x 2 − 8 x − 2 = 2( x + 2) = 2 x + 4 . So out equation simplifies to log 3 (2 x + 4) = 4 . We rewrite this log equation into exponential form, removing the logarithm from the problem. 2 x + 4 = 3 4 = 81 Smith (SHSU) Elementary Functions 2013 9 / 16 Smith (SHSU) Elementary Functions 2013 10 / 16 We can (easily!) solve nice linear equations like 2 x + 4 = 81 and get x = 77 Solving exponential and logarithmic equations 2 . 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 Elementary Functions the universe.) Some numbers are so large it is difficult to even figure out the number of Part 3, Exponential Functions & Logarithms digits. Here is an example. Lecture 3.5b, Solving Equations With Logarithms, continued How many decimal digits are there in the number 2 300 3 100 ? We can answer this question by computing the logarithm of the number Dr. Ken W. Smith base 10 and correctly interpreting the result. If N is a positive integer then the number of decimal digits in 10 N is Sam Houston State University N + 1 since 10 N is written as a one followed by N zeroes. Since the 2013 logarithm base 10 of 10 N 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. Smith (SHSU) Elementary Functions 2013 11 / 16 Smith (SHSU) Elementary Functions 2013 12 / 16
Solving exponential and logarithmic equations Solving exponential and logarithmic equations The largest known prime number. For example, There are an infinite number of primes. This was first proven by Euclid log(2 300 3 100 ) = log(2 300 ) + log(3 100 ) = 300 log 2 + 100 log 3 . around 300 BC! However, we only know, at this time, a finite number of prime numbers. We can approximate log 2 ≈ 0 . 30103 and log 3 ≈ 0 . 47712 to write Large prime numbers play a role in computer science and digital security. As of June 2013, the largest known prime number (according to 300 log 2+100 log 3 ≈ 300(0 . 30103)+100(0 . 47712) = 90 . 309+47 . 712 = 138 . 021 . Wikipedia) is 2 57 , 885 , 161 − 1 . This is a big number! Suppose I want write This tells us that out this big prime number. How many decimal digits would it take? 2 300 3 100 ≈ 10 138 . 021 . Solution. Note that 10 1 = 10 has two decimal digits, 10 2 = 100 has three decimal log 10 (2 57 , 885 , 161 ) = 57885161 · log 10 (2) = 57885161 ln 2 ln 10 digits and in general, if we want the decimal digits of an expression of the ≈ 57885161(0 . 30103) ≈ 17425169 . 76484 . form 10 x , we need to round up . So 10 138 . 021 has 139 decimal digits. This means that 2 57 , 885 , 161 ≈ 10 17425169 . 76484 = (10 0 . 76484 )(10 17425169 ) ≈ We can say more. We can approximate 10 138 . 021 = 10 0 . 021 × 10 138 ≈ 1 . 05 × 10 138 . So 2 300 3 100 begins “105...” 5 . 81887 × 10 17425169 . and continues with another 136 digits!! In other words, 2 57 , 885 , 161 begins with a 5 and is followed by another Smith (SHSU) Elementary Functions 2013 13 / 16 Smith (SHSU) Elementary Functions 2013 14 / 16 17425169 digits, so it has 17,425,170 digits! That’s over 17 million digits! Exponential Functions Exponential Functions I certainly don’t want to try to write out the 17,425,170 digits of 2 57 , 885 , 161 − 1 . In the next presentation we continue to practice applications of logarithms. Indeed, it might be hard to get a computer system to write that out, although you could give WolframAlpha a try. (END) In the solution to this problem on prime numbers, I calculated the number of digits in 2 57 , 885 , 161 . But the prime number we were after is really 2 57 , 885 , 161 − 1 . Is it obvious that subtracting 1 from 2 57 , 885 , 161 won’t change the number of digits? Smith (SHSU) Elementary Functions 2013 15 / 16 Smith (SHSU) Elementary Functions 2013 16 / 16
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