Download - Solving exponential and logarithmic equationskws006/Precalculus/3.5_Solving...Solving exponential and logarithmic equations Example Solve the following exponential equations for x.

Elementary FunctionsPart 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 solveexponential equations.As a general principle, whenever we seek the value of a variable in anequation:

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 mathnotes 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 theexponent, we may take a logarithm of both sides of the equation. In mostcases, 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 7x = 9

Solutions.

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 9ln 7 .


http://tutorial.math.lamar.edu/Classes/Alg/SolveExpEqns.aspx



2 24y+1 − 3y = 0.

Solutions.

2 Rewrite 24y+1 − 3y = 0 as 24y+1 = 3y. Apply ln() to both sides ofthe equation to obtain

ln(24y+1) = ln(3y)

and pull out the exponents

(4y + 1) ln 2 = y ln 3.

Isolate y.4y 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

y = − ln 24 ln 2−ln 3 = ln 2

ln 3−ln 16 .Smith (SHSU) Elementary Functions 2013 5 / 16



3 et+6 = 2.4 5e2z+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 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 soln 5 + 2z + 4 = ln 8

and so2z = ln 8− ln 5− 4.

z = ln 8−ln 5−42 .




5 105x−8 = 8.

Solutions.

5 Apply log() to both sides of 105x−8 = 8 to obtain 5x− 8 = log 8 andso

x = log 8+85 .

(Note that here we are using 10 as the base of our logarithm in thisproblem.)



Some more worked problems.Here are some problems off of an old exam: Solve for x in the followingequations, finding the exact value of x. Then use your calculator toapproximate the value of x to four decimal places.

1 2x = 172 2x = 3x+1

Solution.

1 x = log2(17) =ln 17ln 2 ≈ 4.08746284.

2 Take the natural log of both sides of the equation 2x = 3x+1 to obtain

ln(2x) = ln(3x+1)

Use the exponent property to pull down the exponents and thenisolate 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.

x = ln 3ln 2−ln 3 ≈ −2.70951129



Sometimes our equation explicitly involves logarithms and we need to useproperties of logarithms to get the problem into the correct form where wecan 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 quotient

property to rewrite the lefthand side as log32x2−8x−2 . We may factor 2x2 − 8

as 2(x− 2)(x+ 2) and so (as long as x 6= 2) simplify2x2−8x−2 = 2(x+ 2) = 2x+ 4. So out equation simplifies to

log3(2x+ 4) = 4.

We rewrite this log equation into exponential form, removing thelogarithm from the problem.

2x+ 4 = 34 = 81

We can (easily!) solve nice linear equations like 2x+ 4 = 81 and get

x = 772 .


Exponential Functions

In the next presentation we continue to practice applications of logarithms.

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Elementary FunctionsPart 3, Exponential Functions & Logarithms

Lecture 3.5b, Solving Equations With Logarithms, continued

Dr. Ken W. Smith

Sam Houston State University

2013



Modern scientific computations sometimes involve large numbers (such asthe number of atoms in the galaxy or the number of seconds in the age ofthe universe.)Some numbers are so large it is difficult to even figure out the number ofdigits. Here is an example.How many decimal digits are there in the number 23003100?We can answer this question by computing the logarithm of the numberbase 10 and correctly interpreting the result.

If N is a positive integer then the number of decimal digits in 10N isN + 1 since 10N is written as a one followed by N zeroes. Since thelogarithm base 10 of 10N is just N , we see that the number of decimaldigits in a number is one more than the floor of the logarithm base ten.



For example,

log(23003100) = log(2300) + log(3100) = 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 that23003100 ≈ 10138.021.

Note that 101 = 10 has two decimal digits, 102 = 100 has three decimaldigits and in general, if we want the decimal digits of an expression of theform 10x, we need to round up. So 10138.021 has 139 decimal digits.

We can say more. We can approximate10138.021 = 100.021 × 10138 ≈ 1.05× 10138. So 23003100 begins “105...”and continues with another 136 digits!!



The largest known prime number.There are an infinite number of primes. This was first proven by Euclidaround 300 BC! However, we only know, at this time, a finite number ofprime numbers.

Large prime numbers play a role in computer science and digital security.As of June 2013, the largest known prime number (according toWikipedia) is 257,885,161 − 1. This is a big number! Suppose I want writeout this big prime number. How many decimal digits would it take?Solution.

log10(257,885,161) = 57885161 · log10(2) = 57885161 ln 2

ln 10

≈ 57885161(0.30103) ≈ 17425169.76484.

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 another17425169 digits, so it has 17,425,170 digits! That’s over 17 million digits!Smith (SHSU) Elementary Functions 2013 14 / 16


I certainly don’t want to try to write out the 17,425,170 digits of257,885,161 − 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 numberof digits in 257,885,161. But the prime number we were after is really257,885,161 − 1. Is it obvious that subtracting 1 from 257,885,161 won’tchange the number of digits?



In the next presentation we continue to practice applications of logarithms.

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http://en.wikipedia.org/wiki/Largest_known_prime_number