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Elementary Functions
Part 3, Exponential Functions & Logarithms Lecture 3.5a, Solving Equations With Logarithms
Dr. Ken W. Smith
Sam Houston State University
2013
Smith (SHSU) Elementary Functions 2013 1 / 16
Solving exponential and logarithmic equations
We explore some results involving exponential equations and logarithms.
In this presentation 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 logarithm
Smith (SHSU) Elementary Functions 2013 2 / 16
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 notes on logarithms at Lamar University.)
Example Solve the following exponential equations for x. 1 7 x^ = 9 2 24 y+1^ − 3 y^ = 0. 3 et+6^ = 2. 4 5 e^2 z+4^ − 8 = 0 5 105 x−^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.
Solving exponential and logarithmic equations
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
ln(7x) = ln(9)
and so (by the “exponent” property of logs)
x ln(7) = ln(9)
and so x = ln 9ln 7.
Solving exponential and logarithmic equations
Example Solve the following exponential equations for x. 2 24 y+1^ − 3 y^ = 0.
Solutions. 2 Rewrite 24 y+1^ − 3 y^ = 0 as 24 y+1^ = 3y. Apply ln() to both sides of the equation to obtain ln(2^4 y+1) = ln(3y) and pull out the exponents (4y + 1) ln 2 = y ln 3. Isolate y. 4 y ln 2 − y ln 3 = − ln 2. 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
Smith (SHSU) y^ =^ −Elementary Functions^ 4 ln 2ln 2−ln 3 =^ ln 3ln 2−ln 16. 2013 5 / 16
Solving exponential and logarithmic equations
Example Solve the following exponential equations for x. 3 et+6^ = 2. 4 5 e^2 z+4^ − 8 = 0 Solutions. 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 5 e^2 z+4^ = 8 to obtain ln 5e^2 z+4^ = ln 8 Use the “multiplication” property of logs to rewrite this as ln 5 + ln e^2 z+4^ = ln 8 and so ln 5 + 2z + 4 = ln 8 and so 2 z = ln 8 − ln 5 − 4.
z = ln 8−ln 5 2 −^4.
Smith (SHSU) Elementary Functions 2013 6 / 16
Solving exponential and logarithmic equations
Example Solve the following exponential equations for x. 5 105 x−^8 = 8.
Solutions. 5 Apply log() to both sides of 105 x−^8 = 8 to obtain 5 x − 8 = log 8 and so x = log 8+8 5.
(Note that here we are using 10 as the base of our logarithm in this problem.)
Smith (SHSU) Elementary Functions 2013 7 / 16
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 approximate the value of x to four decimal places. 1 2 x^ = 17 2 2 x^ = 3x+ Solution. 1 x = log 2 (17) = ln 17ln 2 ≈ 4. 08746284. 2 Take the natural log of both sides of the equation 2 x^ = 3x+1^ to obtain ln(2x) = ln(3x+1) Use the exponent property to pull down the exponents and then 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 Smith (SHSU) x(ln 2Elementary Functions − ln 3) = ln 3. 2013 8 / 16
Solving exponential and logarithmic equations
For example,
log(2^3003100 ) = log(2^300 ) + log(3^100 ) = 300 log 2 + 100 log 3.
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 23003100 ≈ 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 10 x, we need to round up. So 10138.^021 has 139 decimal digits.
We can say more. We can approximate 10138.^021 = 10^0.^021 × 10138 ≈ 1. 05 × 10138. So 23003100 begins “105...” and continues with another 136 digits!! Smith (SHSU) Elementary Functions 2013 13 / 16
Solving exponential and logarithmic equations
The largest known prime number. There are an infinite number of primes. This was first proven by Euclid around 300 BC! However, we only know, at this time, a finite number of prime numbers.
Large prime numbers play a role in computer science and digital security. As of June 2013, the largest known prime number (according to Wikipedia) 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. log 10 (2^57 ,^885 ,^161 ) = 57885161 · log 10 (2) = 57885161 (^) ln 10ln 2 ≈ 57885161(0.30103) ≈ 17425169. 76484. This means that 257 ,^885 ,^161 ≈ 1017425169.^76484 = (10^0.^76484 )(10^17425169 ) ≈
- 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 17Smith^ (SHSU)^ Elementary Functions^ million^2013 digits!^ 14 / 16
Exponential Functions
I certainly don’t want to try to write out the 17,425,170 digits of 257 ,^885 ,^161 − 1.
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?
Exponential Functions
In the next presentation we continue to practice applications of logarithms.
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