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Hi,
Okay, I just went over some old stuff this week to prepare for the next vast vast, upcoming test.
I came across a Logarithmic Equation book, well, it's more than 300 pages long, it covers most things from Exponential to Logarithm. I have read about 1/4 of it, and I was stuck.
There are some problems there, but it's like they give out 10 problems, and they only guide you to solve 2, or 3 of them. And the others, you have to do it on your own. >"<
The chapter I was reading was about Using the Mean Value Theorem to Equations. Here's a brief (well, not-so-short, to put it exactly) summary of this chapter, in case anyone needs it . The problems I stuck on are at the end of the post.
---------------------------------
Sumary:
Firstly, the book states the Lagrange's theorem:
If f(x) is continuous on the interval [a; b], and f(x) is differentiable on [a; b], then there will always exist a c on (a; b) such that:
Then, it says that there are 2 kinds of problems:
The first type:: Prove that the equation f(x) = 0 has root(s):
First step is to find F(x), the anti-derivative of f(x), differentiable, and continuous on the interval [a; b], such that: F(b) - F(a) = 0.
Then, there exists an x0 on (a; b), such that:
[tex]F'(x_0) = \frac{F(b) - F(a)}{b - a} = 0 \Rightarrow f(x_0) = 0[/tex]
So that means the equation f(x) = 0 has at least one root on (a; b).
Example:
Given a/3 + b/2 + c = 0, prove that
a22x + b2x + c = 0 always has solution.
Solution:
Let t = 2x ~~> t > 0
Consider F(t) = (a/3)t3 + (b/2)t2 + ct. It's differentiable, and continuous on [tex](0 ; \infty )[/tex]
F'(t) = at2 + bt + c = a22x + b2x + c
F(1) - F(0) = (a/3 + b/2 + c) - 0 = a/3 + b/2 + c = 0
So, the equation F'(t) = 0 (i.e at2 + bt + c = 0) always has solution on he interval (0; 1), hence, the former equation also has solution (Q.E.D)
The second type:: Solve an equation:
First step is to assume that [tex]\alpha[/tex] is the solution to the equation.
Re-arrange the equation into the form f(a) = f(b), from there, find the appropriate f(t), continuous, and differentiable on [a; b].
Then, by using the Mean Value Theorem, we'll solve for [tex]\alpha[/tex].
And the last step is to check the solution.
Example:
Solve:
6x + 2x = 5x + 3x
Solution:
Re-arrange the equation to give:
6x - 5x = 3x - 2x
Assume that [tex]\alpha[/tex] is the solution to the above equation. So, we have:
[tex]6 ^ \alpha - 5 ^ \alpha = 3 ^ \alpha - 2 ^ \alpha[/tex] (1)
Let [tex]f(t) = (t + 1) ^ \alpha - t ^ \alpha, \ \ \ \ \ t > 0[/tex], from (1), we have: f(5) = f(2)
By the Mean Value Theorem, there exists a c on the interval (2; 5), such that:
[tex]f'(c) = \frac{f(5) - f(2)}{5 - 2} = 0 \Rightarrow \alpha (c + 1) ^ {\alpha - 1} - \alpha c ^ {\alpha - 1} = 0[/tex]
[tex]\Rightarrow \alpha \left[ (c + 1) ^ {\alpha - 1} - c ^ {\alpha - 1} \right] = 0 \Rightarrow \left[ \begin{array}{l} \alpha = 0 \\ \alpha = 1 \end{array} \right.[/tex]
So, the two solution to the equation is x = 0, or x = 1. After testing the two solution, we see that they are all valid.
Example:
Solve for x:
3cos(x) - 2cos(x) = cos(x)
Solution:
Re-arrange the equation to give:
3cos(x) -3 cos(x) = 2cos(x) - 2cos(x)
Assume that [tex]\alpha[/tex] is the solution to the equation, we have:
[tex]3 ^ {\cos (\alpha) } - 3 \cos \alpha = 2 ^ {\cos (\alpha)} - 2 \cos \alpha[/tex] (2)
Let [tex]f(t) = t ^ {\cos \alpha} - t \cos \alpha[/tex]
From (2), we have: f(3) = f(2)
Applying the Mean Value Theorem, we have:
There exists a c on (2, 3), such that:
[tex]f'(c) = 0 \Rightarrow \left[c ^ {\cos \alpha} - 1 \right] \cos \alpha = 0 \Rightarrow \left[ \begin{array}{l} \cos \alpha = 0 \\ \cos \alpha = 1 \end{array} \right.[/tex]
[tex]\Rightarrow \left[ \begin{array}{l} \alpha = \frac{\pi}{2} + k \pi \\ \alpha = 2 k' \pi \end{array} \right. , \ \ \ \ \ k, k' \in \mathbb{Z}[/tex]
Testing the two solutions above, we see that they are all valid, so there are 2 solution to the equation:
[tex]\Rightarrow \left[ \begin{array}{l} x = \frac{\pi}{2} + k \pi \\ x = 2 k' \pi \end{array} \right. , \ \ \ \ \ k, k' \in \mathbb{Z}[/tex]
------------------------
Pfff, so, well, it's the summary.
And there are 2 problems I stuck on:
Problem 1:
Given a, b, c in Z+, i.e {1, 2, 3, 4, ...}, and a2 + b2 = c2
Prove that:
ax + bx = cx only has one solution.
Attempt:
a2 + b2 = c2 means that c > a, and c . b.
Divide both sides by cx, to get:
(a/c)x + (b/c)x = 1
The LHS is a decreasing function, hence, if there is root to the equation, it cannot be more than 1.
From the given condition, one can see that x = 2 is the only solution. :)
It seems correct. Except that I haven't used the fact that a, b, and c are in Z+, my solution is valid for, a, b, and c in R+. So, I think I must be missing something here.
Is it me, or the book mistype the Z+ thingy? Or does the book expect me to solve it in another way?
Problem 2:
Find all solution to the equation:
(1 + cos x)(2 + 4cos(x)) = 3 (4cos x)
Well, I have absolutely no idea how to start it. Some body please give me a start.
Thanks very much. :)
Okay, I just went over some old stuff this week to prepare for the next vast vast, upcoming test.
I came across a Logarithmic Equation book, well, it's more than 300 pages long, it covers most things from Exponential to Logarithm. I have read about 1/4 of it, and I was stuck.
There are some problems there, but it's like they give out 10 problems, and they only guide you to solve 2, or 3 of them. And the others, you have to do it on your own. >"<
The chapter I was reading was about Using the Mean Value Theorem to Equations. Here's a brief (well, not-so-short, to put it exactly) summary of this chapter, in case anyone needs it . The problems I stuck on are at the end of the post.
---------------------------------
Sumary:
Firstly, the book states the Lagrange's theorem:
If f(x) is continuous on the interval [a; b], and f(x) is differentiable on [a; b], then there will always exist a c on (a; b) such that:
[tex]f'(c) = \frac{f(b) - f(a)}{b - a}[/tex]
Then, it says that there are 2 kinds of problems:
The first type:: Prove that the equation f(x) = 0 has root(s):
First step is to find F(x), the anti-derivative of f(x), differentiable, and continuous on the interval [a; b], such that: F(b) - F(a) = 0.
Then, there exists an x0 on (a; b), such that:
[tex]F'(x_0) = \frac{F(b) - F(a)}{b - a} = 0 \Rightarrow f(x_0) = 0[/tex]
So that means the equation f(x) = 0 has at least one root on (a; b).
Example:
Given a/3 + b/2 + c = 0, prove that
a22x + b2x + c = 0 always has solution.
Solution:
Let t = 2x ~~> t > 0
Consider F(t) = (a/3)t3 + (b/2)t2 + ct. It's differentiable, and continuous on [tex](0 ; \infty )[/tex]
F'(t) = at2 + bt + c = a22x + b2x + c
F(1) - F(0) = (a/3 + b/2 + c) - 0 = a/3 + b/2 + c = 0
So, the equation F'(t) = 0 (i.e at2 + bt + c = 0) always has solution on he interval (0; 1), hence, the former equation also has solution (Q.E.D)
The second type:: Solve an equation:
First step is to assume that [tex]\alpha[/tex] is the solution to the equation.
Re-arrange the equation into the form f(a) = f(b), from there, find the appropriate f(t), continuous, and differentiable on [a; b].
Then, by using the Mean Value Theorem, we'll solve for [tex]\alpha[/tex].
And the last step is to check the solution.
Example:
Solve:
6x + 2x = 5x + 3x
Solution:
Re-arrange the equation to give:
6x - 5x = 3x - 2x
Assume that [tex]\alpha[/tex] is the solution to the above equation. So, we have:
[tex]6 ^ \alpha - 5 ^ \alpha = 3 ^ \alpha - 2 ^ \alpha[/tex] (1)
Let [tex]f(t) = (t + 1) ^ \alpha - t ^ \alpha, \ \ \ \ \ t > 0[/tex], from (1), we have: f(5) = f(2)
By the Mean Value Theorem, there exists a c on the interval (2; 5), such that:
[tex]f'(c) = \frac{f(5) - f(2)}{5 - 2} = 0 \Rightarrow \alpha (c + 1) ^ {\alpha - 1} - \alpha c ^ {\alpha - 1} = 0[/tex]
[tex]\Rightarrow \alpha \left[ (c + 1) ^ {\alpha - 1} - c ^ {\alpha - 1} \right] = 0 \Rightarrow \left[ \begin{array}{l} \alpha = 0 \\ \alpha = 1 \end{array} \right.[/tex]
So, the two solution to the equation is x = 0, or x = 1. After testing the two solution, we see that they are all valid.
Example:
Solve for x:
3cos(x) - 2cos(x) = cos(x)
Solution:
Re-arrange the equation to give:
3cos(x) -3 cos(x) = 2cos(x) - 2cos(x)
Assume that [tex]\alpha[/tex] is the solution to the equation, we have:
[tex]3 ^ {\cos (\alpha) } - 3 \cos \alpha = 2 ^ {\cos (\alpha)} - 2 \cos \alpha[/tex] (2)
Let [tex]f(t) = t ^ {\cos \alpha} - t \cos \alpha[/tex]
From (2), we have: f(3) = f(2)
Applying the Mean Value Theorem, we have:
There exists a c on (2, 3), such that:
[tex]f'(c) = 0 \Rightarrow \left[c ^ {\cos \alpha} - 1 \right] \cos \alpha = 0 \Rightarrow \left[ \begin{array}{l} \cos \alpha = 0 \\ \cos \alpha = 1 \end{array} \right.[/tex]
[tex]\Rightarrow \left[ \begin{array}{l} \alpha = \frac{\pi}{2} + k \pi \\ \alpha = 2 k' \pi \end{array} \right. , \ \ \ \ \ k, k' \in \mathbb{Z}[/tex]
Testing the two solutions above, we see that they are all valid, so there are 2 solution to the equation:
[tex]\Rightarrow \left[ \begin{array}{l} x = \frac{\pi}{2} + k \pi \\ x = 2 k' \pi \end{array} \right. , \ \ \ \ \ k, k' \in \mathbb{Z}[/tex]
------------------------
Pfff, so, well, it's the summary.
And there are 2 problems I stuck on:
Problem 1:
Given a, b, c in Z+, i.e {1, 2, 3, 4, ...}, and a2 + b2 = c2
Prove that:
ax + bx = cx only has one solution.
Attempt:
a2 + b2 = c2 means that c > a, and c . b.
Divide both sides by cx, to get:
(a/c)x + (b/c)x = 1
The LHS is a decreasing function, hence, if there is root to the equation, it cannot be more than 1.
From the given condition, one can see that x = 2 is the only solution. :)
It seems correct. Except that I haven't used the fact that a, b, and c are in Z+, my solution is valid for, a, b, and c in R+. So, I think I must be missing something here.
Is it me, or the book mistype the Z+ thingy? Or does the book expect me to solve it in another way?
Problem 2:
Find all solution to the equation:
(1 + cos x)(2 + 4cos(x)) = 3 (4cos x)
Well, I have absolutely no idea how to start it. Some body please give me a start.
Thanks very much. :)
Last edited: