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Logarithms/intro to logarithms (how did they the get solution?)

  1. Aug 1, 2012 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations



    3. The attempt at a solution


    24lua8i.jpg


    I have no idea how they came up with 7/2 as a solution...can anyone tell me how? Im not sure how logarithms and square roots work....how did they get the 1/2 exponent over the second 2? What is the train of thought required to solve a question like this??.... for example

    you do this...then you do that...and using that you get 7/2...can someone explain to me? in detail how they got 7/2?

    Thanks :)
     
    Last edited: Aug 1, 2012
  2. jcsd
  3. Aug 1, 2012 #2

    HallsofIvy

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    Re: Logarithms/intro to logarithms

    Your links don't work.
     
  4. Aug 1, 2012 #3
    Re: Logarithms/intro to logarithms

    Try again...i re did them...and they seem to work now.
     
  5. Aug 1, 2012 #4

    HallsofIvy

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    Looks like magic!

    Do you know the "properties of exponents"? [itex]a^xa^y= a^{x+y}[/itex] and [itex](a^x)^y= a^{xy}[/itex]

    The logarithm is the opposite of the exponential: [itex]log_a(a^x)= x[/itex] and [itex]a^{log_a(x)}= x[/itex] (the technical term is "inverse") so logarithms have the opposite properties: [itex]log_a(xy)= log_a(x)+ log_a(y)[/itex] and [itex]log_a(x^y)= y log_a(x)[/itex]

    So, for example, [itex]log_3(27\sqrt{3})= log_3(3^3(3^{1/2})= log_3(3^{3+ 1/2})= log_3(3^{7/2})= 7/2[/itex]
    That, as well as, [itex]log_2(8\sqrt{2})= log_2(2^3(2^{1/2})= log_2(2^{3+ 1/2})= log_2(2^{7/2})= 7/2[/itex]
    are both applications of [itex]log_a(a^x)= x[/itex]. The "logarithm" and "exponential" are inverse functions so they "cancel".
     
  6. Aug 1, 2012 #5
    A notation that is used for square roots is to raise the number to the one-half power.

    So the square root of two is equivalent to ##2^{\frac{1}{2}}##.
    On a similar note, the cube root of two is equivalent to ##2^{\frac{1}{3}}##. (this isn't relevant for this problem, but I feel like it helps you understand the rule a little better.
     
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