Download 3.5 Log Equations (slides 4 to 1)

Solving exponential and logarithmic equations
We explore some results involving exponential equations and logarithms.
Elementary Functions
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:
Part 3, Exponential Functions & Logarithms
Lecture 3.5a, Solving Equations With Logarithms
Dr. Ken W. Smith
Sam Houston State University
If the variable appears as an exponent, we should think about using logarithm
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Elementary Functions
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
notes on logarithms at Lamar University.)
Example Solve the following exponential equations for x.
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7x = 9
Example Solve the following exponential equations for x.
1
7x = 9
2
24y+1 − 3y = 0.
3
et+6 = 2.
4 5e2z+4
5
Solutions.
1
Apply ln() to both sides of 7x = 9 to obtain
ln(7x ) = ln(9)
−8=0
and so (by the “exponent” property of logs)
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.
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x ln(7) = ln(9)
and so x =
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ln 9
ln 7 .
Elementary Functions
Solving exponential and logarithmic equations
Solving exponential and logarithmic equations
Example Solve the following exponential equations for x.
2 24y+1 − 3y = 0.
Example Solve the following exponential equations for x.
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
Solutions.
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
ln 5e2z+4 = ln 8
(4y + 1) ln 2 = y ln 3.
Use the “multiplication” property of logs to rewrite this as
Isolate y.
ln 5 + ln e2z+4 = ln 8
4y ln 2 − y ln 3 = − ln 2.
and so
Factor out y:
ln 5 + 2z + 4 = ln 8
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
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y=
ln 2
ln 2
−Elementary
4 ln 2−lnFunctions
3 = ln 3−ln 16 .
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and so
2z = ln 8 − ln 5 − 4.
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Elementary Functions
z=
Solving exponential and logarithmic equations
105x−8 = 8.
Solutions.
5
Apply log() to both sides of 105x−8 = 8 to obtain 5x − 8 = log 8 and
so
x = log 58+8 .
(Note that here we are using 10 as the base of our logarithm in this
problem.)
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.
Solving exponential and logarithmic equations
Example Solve the following exponential equations for x.
5
ln 8−ln 5−4
2
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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.
x
1 2 = 17
x
x+1
2 2 =3
Solution.
ln 17
1 x = log2 (17) = ln 2 ≈ 4.08746284.
x
x+1
2 Take the natural log of both sides of the equation 2 = 3
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
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x(ln
2 − Functions
ln 3) = ln 3.
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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.
log3 (2x2 − 8) − log3 (x − 2) = 4.
(END)
Solution. We solve log3 (2x2 − 8) − log3 (x − 2) = 4 by using our quotient
2 −8
. We may factor 2x2 − 8
property to rewrite the lefthand side as log3 2xx−2
as 2(x − 2)(x + 2) and so (as long as x 6= 2) simplify
2x2 −8
x−2 = 2(x + 2) = 2x + 4. So out equation simplifies to
log3 (2x + 4) = 4.
We rewrite this log equation into exponential form, removing the
logarithm from the problem.
2x + 4 = 34 = 81
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We can (easily!) solve nice linear equations like 2x + 4 = 81 and get
x=
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77
2 .
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Solving exponential and logarithmic equations
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.
Elementary Functions
Part 3, Exponential Functions & Logarithms
Lecture 3.5b, Solving Equations With Logarithms, continued
Dr. Ken W. Smith
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.
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Solving exponential and logarithmic equations
Solving exponential and logarithmic equations
For example,
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.
log(2300 3100 ) = log(2300 ) + log(3100 ) = 300 log 2 + 100 log 3.
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 257,885,161 − 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?
2300 3100 ≈ 10138.021 .
Solution.
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.
2
log10 (257,885,161 ) = 57885161 · log10 (2) = 57885161 lnln10
≈ 57885161(0.30103) ≈ 17425169.76484.
This means that
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 digits!!
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257,885,161 ≈ 1017425169.76484 = (100.76484 )(1017425169 ) ≈
5.81887 × 1017425169 .
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Exponential Functions
In other words, 257,885,161 begins with a 5 and is followed by another
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17425169
digits, so it has 17,425,170
digits! That’s over 17 million
digits!
Exponential Functions
I certainly don’t want to try to write out the 17,425,170 digits of
257,885,161 − 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 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?
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