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Some Maths Problems

  1. Jun 6, 2005 #1

    I need some help with some maths problems. Actually I am sorta of confused.

    My Questions:
    1. Rewrite without rational exponents, and simplify, if possible.
    a. y^(1/7)...... I am so confused what it is about. I mean I can't do anything with the expressions, can I?
    b. (x^2.y^2)^(1/3)...... kinda of similar to above one but wat am I supposed to do with expression....

    2. Rewrite with positive exponents.
    a. y^(-1/7)........ Am I supposed to just change the sign of power to +ve??

    3. Use rational exponents to simplify. Write the answer in radical notation if appropriate.
    a. (a^2)^(6)......... :confused:

    4. Find domain of f(x)= (x+5)^(1/2). (x+7)^(-1/2) hints badly needed...

    I have test tomm so pls reply ASAP. Don't solve it right away but provide me with hints pls.

  2. jcsd
  3. Jun 6, 2005 #2
    Sounds like you have difficulty with the language used which will keep you from even starting the problems really.

    1a. y^(1/7) is the same as writing the 7th root of y so.. write the 7th root of y a different way.
    b. same thing basically as a
    2. y^-x = 1/(y^x) right?
    3. more power rules. How would you combine the 2 and the 6?
    4. Not sure what the "." stands for, but assuming it's multiplication.. simply find the values of x that will result in an imaginary answer. for example, f(-6) is not in the domain.
  4. Jun 6, 2005 #3
    Well, Im not too sure exactly what the question is asking, and what context it is in but I would say that for the first question it is asking you to write it in surd form.

    So for 1) a)

    [tex]y^{\frac{1}{7}} = \sqrt[7]{y}[/tex]


    [tex](x^2y^2)^{\frac{1}{3}} = x^{\frac{2}{3}}y^{\frac{2}{3}} = (xy)^{\frac{2}{3}} = \sqrt[3]{xy}^2[/tex]


    [tex]y^{-1/7} = \frac{1}{y^{1/7}}[/tex]

    3) I dont exactly get the question? There is not much else you can do with this type of question.

    [tex](a^2)^6 = a^{12}[/tex]

    4) We have

    [tex]f(x) = \frac{\sqrt{x+5}}{\sqrt{x+7}}[/tex]

    As x approaches -7 from [itex]\infty^-[/itex] it hits an asymptote. So [itex]x\neq -7[/itex] otherwise the denominator is zero. Also, [itex]x \notin (-7,-5)[/itex]. x cannot exist in this interval because

    [tex]f(-7) = \frac{\sqrt{-2}}{\sqrt{0}}[/tex] does not exist

    [tex]f(-6) = \frac{\sqrt{-1}}{\sqrt{1}} = i \notin \mathbb{R}[/tex]

    [tex]f(-5.5) = \frac{\sqrt{-1/2}}{\sqrt{1.5}} = \frac{i}{3}\sqrt{3} \notin \mathbb{R}[/tex]

    [tex]f(-5) = 0[/tex]

    Hence the domain of [itex]f(x)[/itex] is

    [tex] f(x) = (-\infty, -7) \cup (-5,\infty) [/tex]


    [tex] f(x) \backslash (-7,-5)[/tex]
    Last edited: Jun 6, 2005
  5. Jun 6, 2005 #4
    Thanx guys

    my bad vsage..... misunderstood it I guess....

    thanx for help guys, vsage and oxymoron..... :smile:

    and yeah the third Q is:

    Use rational exponents to simplify. Write the answer in radical notation if appropriate.


    so pls tell me what I am supposed to do ........ thanx :smile:
    Last edited by a moderator: Jun 6, 2005
  6. Jun 7, 2005 #5
    Well, to write in radical notation is to write in surd form (with square roots).


    [tex] a^{\frac{2}{6}} = \sqrt[6]{a}^2[/tex]

    By the way, Im using the surd formula:

    [tex]x^{\frac{y}{z}} = \sqrt[z]{x}^y[/tex]
  7. Jun 7, 2005 #6
    Also note that with your 4th question, when regarding domains of functions, you must note what field you are working in.

    When I did the question I assumed you were working with real numbers.

    But if you were working with complex numbers, obviously the function is defined for all [itex]z\in\mathbb{C}\backslash\{-7\}[/tex].
  8. Jun 7, 2005 #7
    Thanx oxy and vsage that helped me.....

    I did my test hope to get an A :smile:

    Just one more Q is:

    log(x-1)/log(3)= [log(x-2)/log3]-[log(x)/log(3)]

    evaluate or simplify the above expression.....

    thanx very much appreciated......
  9. Jun 8, 2005 #8
    in your last post Bin Qasim, you say evaluate or simplify. Well, you cant evalute what you have written but you can simplify. This is just a matter of remembering your logarithm laws...

    [tex]\frac{\log (x-1)}{\log 3} = \frac{\log (x-2)}{\log 3} - \frac{\log x}{\log 3}[/tex]

    [tex]\frac{\log (x-1)}{\log 3} = \frac{\log (x-2) - \log x}{\log 3}[/tex]

    However, this should not be an equality since the LHS never equals the RHS. Are you sure there isnt more to this question?
  10. Jun 14, 2005 #9
    U right oxy, it was to solve for x i forgot that

    and one more thing, I go an A in the taht test, I owe u guys for helping me :smile:
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