QUESTION 3:Calculate the value of log base 3 of 81SOLUTION:Answer is 4.

[petuniac]

QUESTION 1:

Solve log x = 2 Cos x

SOLUTION:

Answer is 1.48, 5.07

What I have done so far:
let y = log x
y = 2 cos x
i graphed this and found the x-intercepts to be 1.57 and 4.71
not sure where to go from here...

QUESTION 2:

Determine the phase shift of the function f(x) = cos (1/3 x - pi)

SOLUTION:

from the question is "pi" not the phase shift?? the answer says the phase shift is 3pi, but I am not sure this is right.

 

[GregA]

Until someone wiser comes along with more advice...

For the second question note that $$cos[\frac{x}{3}-\pi]= cos(\frac{1}{3}[x-3\pi])$$

For the first (and you may want to re-check your plots btw)...though I haven't yet done any work on *solving* these types of equations yet you can, by using Newtons Method, home in on the solutions. This however requires calculus and so it may be beyond the scope of what the question requires, but if you're interested...

noting that log₁₀x = (1/ln10)(lnx) you can differentiate log₁₀x - 2cos x and then plug decent guesses into $x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}$ and keep using your answer as the next guess.(I use log₁₀x because this base fits the answer, ln x (I believe sometimes referred to as log x) doesn't)

(additionally is the domain of x meant to be [0,2pi]?... because only if x > 100 is there no potential for 2cos x = log₁₀x)

 

[arunbg]

For the first question, you need to find the x coordinates of the points of intersection of the two curves if you wish to solve the question graphically.
It is only at these points that the value of the two functions become equal for the same value of x , which is then the solution .

The second question is ambiguous.
Usually, phase shift is defined with respect to some given function, which has the same period as the function in the problem .
Please give the question in its entire form.

Arun

 

[petuniac]

Hi Arunbg... how could I solve question 1 non-graphically?? The domain for this question is 0 to 2pi. Also, that is the complete question for Question 2.

 

[berkeman]

QUESTION 1:

Solve log x = 2 Cos x

SOLUTION:

Answer is 1.48, 5.07

The second solution by the book is correct, but log(1.48) = 0.170 and 2cos(1.48) = 0.181. :uhh:

 

[petuniac]

p.s. i was able to get the correct answers graphing.

 

[berkeman]

p.s. i was able to get the correct answers graphing.

i graphed this and found the x-intercepts to be 1.57 and 4.71

log(1.57) = 0.196, 2cos(1.57) = 0.002

log(4.71) = 0.673, 2cos(4.71) = -0.005

 

[petuniac]

check your calculator mode.. it should be in radians.

log(1.48) = .17 and 2Cos(1.48) = 0.18

log(5.07) = .70 and 2Cos(5.07) = .70

Therefore, the solutions are correct.

 

[berkeman]

check your calculator mode.. it should be in radians.

log(1.48) = .17 and 2Cos(1.48) = 0.18

Therefore, the solutions are correct.

Yeah, I switched to radians for my calcs as well. I just don't consider 0.170 = 0.181

But I just did a more careful check, and the answer is closer to 1.485. That reduces the error to almost nothing. I guess the book was happy with giving the answer to only 2 decimal places, even though that gives a 6% error. In the real world, a 6% error can be a very bad thing...

 

[arunbg]

Unless there is some alternate definition for phase shift, I don't think there is an answer to question 2.
I can only safely say that the period of the function is $3\pi$ .
Perhaps that is what the question is asking ?

Arun

 

[maverick280857]

Unless there is some alternate definition for phase shift, I don't think there is an answer to question 2.
I can only safely say that the period of the function is $3\pi$ .
Perhaps that is what the question is asking ?

Arun

There are different interpretations of phase. Some authors (Halliday/Resnick) consider the argument of the sinusoidal function as the phase which means $(x/3 - \pi)$ is the phase. Some consider the constant part of the argument that is not a function of the independent variable x as the phase, so then $\pi$ is the phase.

A geometric interpretation is that the graph of $\cos (x/3)$ when shifted by $3\pi$ units along the positive x direction gives the graph of f(x). Your books seems to consider phase shift as the amount of displacement that must be given to the first function to construct f(x).

The period is $2\pi/(1/3) = 6\pi$ and not $3\pi$ but that is not being asked here.

 

[arunbg]

A geometric interpretation is that the graph of when shifted by units along the positive x direction gives the graph of f(x). Your books seems to consider phase shift as the amount of displacement that must be given to the first function to construct f(x).

This was what I was trying to allude to in my previous post .
The function considered as the standard(from which phase diff. is calculated) here is probably cos(x/3), which is not given in the question.

The period is $2\pi/(1/3) = 6\pi$ and not $3\pi$
but that is not being asked here.

Thanks for the correction .