# Exponential to Logarithm

1. Jun 18, 2007

### VietDao29

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. :surprised 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:
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$​

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:
$$F'(x_0) = \frac{F(b) - F(a)}{b - a} = 0 \Rightarrow f(x_0) = 0$$
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 $$(0 ; \infty )$$

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 $$\alpha$$ 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 $$\alpha$$.

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 $$\alpha$$ is the solution to the above equation. So, we have:

$$6 ^ \alpha - 5 ^ \alpha = 3 ^ \alpha - 2 ^ \alpha$$ (1)

Let $$f(t) = (t + 1) ^ \alpha - t ^ \alpha, \ \ \ \ \ t > 0$$, from (1), we have: f(5) = f(2)

By the Mean Value Theorem, there exists a c on the interval (2; 5), such that:

$$f'(c) = \frac{f(5) - f(2)}{5 - 2} = 0 \Rightarrow \alpha (c + 1) ^ {\alpha - 1} - \alpha c ^ {\alpha - 1} = 0$$

$$\Rightarrow \alpha \left[ (c + 1) ^ {\alpha - 1} - c ^ {\alpha - 1} \right] = 0 \Rightarrow \left[ \begin{array}{l} \alpha = 0 \\ \alpha = 1 \end{array} \right.$$

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 $$\alpha$$ is the solution to the equation, we have:
$$3 ^ {\cos (\alpha) } - 3 \cos \alpha = 2 ^ {\cos (\alpha)} - 2 \cos \alpha$$ (2)

Let $$f(t) = t ^ {\cos \alpha} - t \cos \alpha$$

From (2), we have: f(3) = f(2)

Applying the Mean Value Theorem, we have:

There exists a c on (2, 3), such that:

$$f'(c) = 0 \Rightarrow \left[c ^ {\cos \alpha} - 1 \right] \cos \alpha = 0 \Rightarrow \left[ \begin{array}{l} \cos \alpha = 0 \\ \cos \alpha = 1 \end{array} \right.$$

$$\Rightarrow \left[ \begin{array}{l} \alpha = \frac{\pi}{2} + k \pi \\ \alpha = 2 k' \pi \end{array} \right. , \ \ \ \ \ k, k' \in \mathbb{Z}$$

Testing the two solutions above, we see that they are all valid, so there are 2 solution to the equation:

$$\Rightarrow \left[ \begin{array}{l} x = \frac{\pi}{2} + k \pi \\ x = 2 k' \pi \end{array} \right. , \ \ \ \ \ k, k' \in \mathbb{Z}$$

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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: Jun 18, 2007
2. Jun 18, 2007

### happyg1

Hey,
Just looking at it before I tried anything, I see that cos x=1 is a solution.
So anywhere cos(x)=0 is a solution.
CC

3. Jun 18, 2007

### VietDao29

Yeah, even cos(x) = 1/2 is a solution, I got it when using Trial-and-Error method frantically. But, well, I cannot factor this expression, or re-arrange it to any desirable form.

4. Jun 18, 2007

### happyg1

Allright,
So I multiplied it out and I got:(letting cos(x)=a for easiness)
$$2+2a+a4^a+4^a=3(4^a)$$
Then I used the following:
2=4-2
2a=4a-2a
$$2(4^a)=4(4^a)-2(2^a)$$
$$a4^a=2a4^a-a4^a$$
which gives:
$$4+4a+2a4^a-4(4^a)=2+2a+a4^a-2(2^a)$$
So I think that works.
CC

5. Jun 19, 2007

### VietDao29

Yeah, yeah, it works. Thanks a lot, mate.

Btw, I have one question. How can you split it into that form, is there anything that suggest you to do that, or you get it by chance? I don't think I can think of that re-arrangement though.

6. Jun 19, 2007

### happyg1

I just looked at the form of the terms and realizing that I needed the same types of terms on both sides of the equation, just divided each one up. After I multiplied it out and cancelled the one $$4^a$$ term, I saw that the 4's and 2's for the coefficients would probably work. With just a little tinkering the thing fell out.

I didn't solve it past that. What did you get for solutions? Was the cos(x)=1 and cos(x)=1/2 all there were?
CC

Last edited: Jun 19, 2007