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  1. #1
    Newbie zaga04's Avatar
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    Questions

    Wow, I can't believe I've forgotten how to do these, but I have. Any help would be appreciated.

    1) Simplify 7root(x^9)/5root(x^6)


    2) Solve the equation both algebraically and graphically (I've done it the graphical way, cannot seem to figure out how to do it algebraically)
    abs(4x-3)=5radical(x+4)



    3) Rewrite the expression logbase5(x+3) into an equivalent expression using only natural logarithms.


    4) Describe the transformations that can be used to transform the graph of log(x) to a graph of f(x) = 4log(x+2)-3.


    5) Solve the equation 2sin^2(x)cos(x)=cos(x) algebraically

    6) Find all the exact solutions to 2sin^2(x)+3sin(x)-2=0 on the interval [0,2pi)


    7) Use a graphing calculator to solve the following for x. e^(2x)=3x^2

    I graphed it in my graphing calculator, but I am not sure if you're trying to find the intersection of the two graphs or what? I found the intersection and plugged it into x, but they are not equal, so I am not sure what to do.

    8) State the smallest interval (
    0 ≤theta ≤k) that gives the complete graph of the polar equation r=4cos(5theta)


    By the way, I am not asking you to solve these outright, but just help me. These problems are from PreCalculus, but I just forget how to solve them. Thank you for your time!
    Last edited by zaga04; August 21st 2009 at 09:51 AM.
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  2. #2
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    A different approach

    Hello zaga04

    Welcome to Math Help Forum!
    Quote Originally Posted by zaga04 View Post
    Wow, I can't believe I've forgotten how to do these, but I have. Any help would be appreciated.

    1) Simplify 7root(x^9)/5root(x^6)


    2) Solve the equation both algebraically and graphically (I've done it the graphical way, cannot seem to figure out how to do it algebraically)
    abs(4x-3)=5radical(x+4)



    3) Rewrite the expression logbase5(x+3) into an equivalent expression using only natural logarithms.


    4) Describe the transformations that can be used to transform the graph of log(x) to a graph of f(x) = 4log(x+2)-3.


    5) Solve the equation 2sin^2(x)cos(x)=cos(x) algebraically

    6) Find all the exact solutions to 2sin^2(x)+3sin(x)-2=0 on the interval [0,2pi)


    7) Use a graphing calculator to solve the following for x. e^(2x)=3x^2

    I graphed it in my graphing calculator, but I am not sure if you're trying to find the intersection of the two graphs or what? I found the intersection and plugged it into x, but they are not equal, so I am not sure what to do.

    8) State the smallest interval (
    0 ≤theta ≤k) that gives the complete graph of the polar equation r=4cos(5theta)


    By the way, I am not asking you to solve these outright, but just help me. These problems are from PreCalculus, but I just forget how to solve them. Thank you for your time!
    May I suggest that you'll be more likely to get help if you (a) post one question at a time (see Rule 14); (b) show us that you've made some sort of attempt, so that we can help you at the specific point that you're having a problem with.

    Having said that, here are some suggestions for the first couple of questions.

    (1) I take it that you mean: Simplify \frac{\sqrt[7]{x^9}}{\sqrt[5]{x^6}}.

    You need to use two rules of indices:

    • \sqrt[a]{x^b} = x^{\frac{b}{a}}


    • \frac{x^a}{x^b}=x^{a-b}

    So \sqrt[7]{x^9}=x^{\frac{9}{7}}, and \sqrt[5]{x^6}=... ?

    Then you'll need to subtract the second power of x from the first.

    (2) |4x-3| = 5 \sqrt{x+4}

    \Rightarrow (4x-3)^2 = 25(x+4)

    \Rightarrow 16x^2 - 24x +9 = 25x +100

    \Rightarrow ... ?

    Solve the quadratic equation. (I don't think it factorises, so you'll have to use the formula.)

    Grandad
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  3. #3
    Newbie zaga04's Avatar
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    Sorry about posting so many questions, thought it would be easier than posting so many threads. Thanks for helping me out for the first two problems, make sense. I'll come back with the work I've done for the others.
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  4. #4
    Newbie zaga04's Avatar
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    So I think I got #3, log5(x+3) converted to ln would = ln(x+3)/ln5

    Would #4 be like the graph of 4log(x+2)-3 is vertically stretched by 4, moved horizontally 2 units to the left, and moved vertically down 3 units? I am not sure if this is what they are asking.

    I am not sure how to start #5, #6 or #8 at all. #8 I graph it on my calculator, but I'm not sure what the smallest interval would be that would give the complete graph of the polar equation.

    #7 I already explained what I did, I graphed it on my graphing calculator, but I am not sure how you find x from the graphing calculator? Do you find the intersection of the two graphs e^(2x) and 3x^2?

    That's the thought process I've put into these problems.
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by zaga04 View Post
    So I think I got #3, log5(x+3) converted to ln would = ln(x+3)/ln5

    Would #4 be like the graph of 4log(x+2)-3 is vertically stretched by 4, moved horizontally 2 units to the left, and moved vertically down 3 units? I am not sure if this is what they are asking.

    I am not sure how to start #5, #6 or #8 at all. #8 I graph it on my calculator, but I'm not sure what the smallest interval would be that would give the complete graph of the polar equation.

    #7 I already explained what I did, I graphed it on my graphing calculator, but I am not sure how you find x from the graphing calculator? Do you find the intersection of the two graphs e^(2x) and 3x^2?

    That's the thought process I've put into these problems.
    Q3: Yes, those two expressions are equal (change of base rule)

    Q4: It would be assuming the original graph is log(x)

    Q5: Take cos(x) from both sides

    2sin^2(x)cos(x) - cos(x) = 2sin^2(x)(cos(x)-1) = 0

    and if ab=0 either a=0 or b=0. Normally this kind of question will have limits on solutions because there are an infinite number.

    Q6: Let u = sin(x) and solve the quadratic

    Q7: Yep that's what you do

    <br />
e^{2x} = 3x^2 cannot be solved algebraically hence the calculator

    Q8: It's asking how much space you need to view a complete wave of the equation. Remember cos(ax) is the graph of cos(x) squashed by a factor of a and that cos(x) is periodic over 2pi
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  6. #6
    Newbie zaga04's Avatar
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    Thank you so much
    For #5, I understand the process, but what kind of answer would you get?

    For #6, I got pi/6 and 5pi/6.

    For #7, I got the intersection to be x=-.39.

    For #8, I got the interval to be (0≤theta≤pi)

    Are the ones I did correct?
    Last edited by zaga04; August 21st 2009 at 12:21 PM.
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  7. #7
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by zaga04 View Post
    Thank you so much
    For #5, I understand the process, but what kind of answer would you get?

    For #6, I got pi/6 and 5pi/6.

    For #7, I got the intersection to be x=-.39.

    For #8, I got the interval to be (0≤theta≤pi)

    Are the ones I did correct?
    5. sin^2(x) = 0 or cos(x) = 1

    x = k \pi \: ,\, k\, \in \, \mathbb{Z}

    In this case sin^2(x) is 0 when cos(x) is also 0.

    6,7. Put the values you found back into the equation and see if it works out to be true

    8. See grandad's answer below mine
    Last edited by e^(i*pi); August 21st 2009 at 01:57 PM.
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  8. #8
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    Polar equation sketch

    Hello zaga04
    Quote Originally Posted by zaga04 View Post
    ...
    8) State the smallest interval (
    0 ≤theta ≤k) that gives the complete graph of the polar equation r=4cos(5theta)
    To sketch the polar graph, note that the important values of cosine occur at intervals of \pi/2:

    \cos 0 = 1,\,\cos(\pi/2) = 0,\, \cos(\pi) = -1, \,\cos(3\pi/2)=0,\,\cos(2\pi) = 1, ...

    These will occur when \theta = 0, \pi/10,\, 2\pi/10,\, 3\pi/10,...

    So sketch a diagram with radial lines every \pi/10; i.e. 18^o from 0^o to 180^o.

    Mark the point (4,0). The curve starts here and comes in to the origin with the line \theta = \pi/10 as a tangent. Then it continues through negative values of r (i.e. out the other side into the third quadrant) going out to (-4, \pi/5); comes back to the origin tangential to \theta = 3\pi/10; etc.

    See the attached sketch.

    You'll find that you need 0 \le \theta \le \pi to get the complete curve.

    Grandad
    Attached Thumbnails Attached Thumbnails Questions-untitled.jpg  
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  9. #9
    Newbie zaga04's Avatar
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    Thanks Grandad

    Also I was trying to rework #5, would this be right?

    2sin^2(x)cos(x)=cos(x)
    2sin^2(x)=1
    sin^2(x)=\frac{1}{2}
    sin(x)=\sqrt\frac{1}{2}
    x=\arcsin\sqrt\frac{1}{2}
    x=\frac{\pi}{4}
    Last edited by zaga04; August 21st 2009 at 04:57 PM.
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  10. #10
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    The two great commandments

    Hello zaga04
    Quote Originally Posted by zaga04 View Post
    Thanks Grandad

    Also I was trying to rework #5, would this be right?

    2sin^2(x)cos(x)=cos(x)
    2sin^2(x)=1
    sin^2(x)=\frac{1}{2}
    sin(x)=\sqrt\frac{1}{2}
    x=\arcsin\sqrt\frac{1}{2}
    x=\frac{\pi}{4}
    There are two great commandments of mathematics. The first is this:

    • Thou shalt not divide by zero.

    and the second is:

    • When thou takest a square root, forget not thy plus or minus sign.

    And I'm afraid you've broken them both!

    You've got to be careful you don't divide both sides of an equation by something that might have the value zero. If you do, you might miss possible solutions.

    In 2\sin^2x\cos x = \cos x, where there's a factor of \cos x on both sides, note that \cos x = 0 is a possible solution - it makes both sides have an equal value, 0. So before you divide by \cos x, say that \cos x =0 or ... then do the division. So you get:

     2\sin^2x\cos x = \cos x

     \Rightarrow \cos x = 0 or 2\sin^2x = 1

    The first possibility gives us x = \frac{\pi}{2}, \frac{3\pi}{2}, ...

    And the second: \sin^2x = \frac12

    So don't forget the second commandment:

    \sin x = \pm\frac{1}{\sqrt2}

    So you not only get x = \frac{\pi}{4},\frac{3\pi}{4}, \frac{9\pi}{4}, ... corresponding to \sin x = +\frac{1}{\sqrt2}, but also the odd multiples of \frac{\pi}{4} in between:

    x=\frac{5\pi}{4}, \frac{7\pi}{4}, ...

    which come from \sin x = -\frac{1}{\sqrt2}

    So don't forget to obey the commandments!

    Grandad
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  11. #11
    Newbie zaga04's Avatar
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    Thank you again Grandad and as you can tell I am not very good at math. I was just trying to do it by myself, but thank you for correcting me and telling me the right way.

    So the dots at the end of each answer would correspond that you have to keep on adding multiples of \frac{\pi}{4}? Or as the answer on my sheet can I just leave it as ... at the end?
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  12. #12
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    Hello zaga04
    Quote Originally Posted by zaga04 View Post
    Thank you again Grandad and as you can tell I am not very good at math. I was just trying to do it by myself, but thank you for correcting me and telling me the right way.

    So the dots at the end of each answer would correspond that you have to keep on adding multiples of \frac{\pi}{4}? Or as the answer on my sheet can I just leave it as ... at the end?
    Yes, the three dots symbol, ..., is called an ellipsis (not to be confused with an ellipse, which is a squashed circle), and it indicates that things have been left out.

    You can use it to indicate a finite sequence, like 2, 4, 6, ..., 10, or an infinite one, like \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, ... which continues without ending.

    Grandad
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