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Homework Help: Where do the exponents go?

  1. May 19, 2005 #1
    Could someone please check my work?

    Write as a single logarithm:

    [tex]\frac{1}{4}[2(log_2 x + 3 log_2 y) - 3 log_2 z][/tex]

    I got [tex]log_2(\frac{x^2y^6}{z^3})^\frac{1}{4}[/tex]

    [tex]\frac{3}{5}[\frac{1}{2}(log_2 x + 3 log_2 y) - 2(log_2 x - 4 log_2 y)][/tex]

    I got [tex]log_2 (\frac{x^\frac{1}{2}y^{1.5}}{x^2} \cdot y^8)^\frac{3}{5}[/tex]

    I'd also appreciate some hints on the following:

    [tex]a^2 + b^2 = 2ab[/tex]
    prove
    [tex]log (\frac{a + b}{2} = \frac{1}{2}(log a + log b)[/tex]

    The part I really don't get is where the exponents go...
     
    Last edited: May 19, 2005
  2. jcsd
  3. May 19, 2005 #2

    JFo

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    first part is perfect although you could get rid of the 1/4 power by looking at the third equation I gave you for the second problem

    the second could be simplified further by noting that

    [tex] \frac{a^m}{a^n} = a^{m-n} [/tex]

    [tex] a^m * a^n = a^{m+n} [/tex]

    [tex] {(a^m)}^n = a^{mn} [/tex]

    for the third, note that

    [tex] (a + b)^2 = a^2 +2ab + b^2 [/tex]

    can you do the rest?
     
    Last edited: May 19, 2005
  4. May 19, 2005 #3
    Work looks fine.


    Suppose that
    [tex]a^2 + b^2 = 2ab[/tex]
    <==>
    [tex]a^2 - 2ab + b^2 = 0[/tex]
    <==>
    [tex](a - b)^2 = 0[/tex]
    then what can we say about a and b?
     
  5. May 19, 2005 #4

    JFo

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    the above hint will work, but in a kind of backwards way... its easier (and more direct) to come up with an expression for (a+b)/2 and use the properties of logs
     
  6. May 19, 2005 #5
    And by above you're referring to your own? If that is the case, I wouldn't agree with you.
     
  7. May 19, 2005 #6

    JFo

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    no I was refering to your hint
     
  8. May 19, 2005 #7
    Ok, but using my hint one don't have to use any properties of logarithms..
     
  9. May 19, 2005 #8

    JFo

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    true, but going that route bypasses what I think the excercise is meant to do, considering the context of the first two questions.

    but I'll let the OP decide what method to use.
     
  10. May 19, 2005 #9

    dextercioby

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    [tex] a^{2}+b^{2}=2 ab [/tex]

    Add [itex] 2ab [/itex] to it:

    [tex] (a+b)^{2}=4ab [/tex]

    You can get

    [tex] \frac{a+b}{2}=\pm (ab)^{\frac{1}{2} [/tex]

    Choose the "+" sign in the RHS,so you won't get imaginary numbers.

    Daniel.
     
  11. May 19, 2005 #10

    JFo

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    yeah...what dextercioby did was what I had in mind
     
  12. May 19, 2005 #11

    I fixed up my first two answer, but I don't understand why you add a^2 and b^2 to the right side for the third. I think I see how I can do it now, though:

    [tex](a + b)^2 = 2ab[/tex]
    [tex]\frac{(a + b)^2}{2} = ab[/tex]
    [tex]2 log (\frac{a + b}{2}) = log a + log b[/tex]
    [tex]log (\frac{a + b}{2}) = \frac{1}{2}(log a + log b)[/tex]

    dextercioby: I see where you're going, but can you explain more how you got the last step?
     
  13. May 19, 2005 #12

    dextercioby

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    By taking sq.root from both sides of the equation ??That's how the "+-" got there.

    And your post is completely wrong.

    Daniel.
     
  14. May 19, 2005 #13

    JFo

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    [tex](a + b)^2 = a^2 + b^2 + 2ab [/tex]

    the above equation is true for any 2 real numbers a and b. [to verify, just multiply out (a+b)(a+b)]

    They told you that [itex] a^2 + b^2 = 2ab [/itex], substituting this into the first equation you get

    [tex](a + b)^2 = 4ab [/tex]

    taking the square root of both sides, you get

    [tex] (a+b) = 2(ab)^{\frac{1}{2}} [/tex]

    (we take the positive root, because the log of a negative number is not defined)

    divide both sides by two and

    [tex] \frac{a+b}{2} = (ab)^{\frac{1}{2}} [/tex]

    take the log of both sides and..... (fill in rest)
     
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