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If Logb(x) = 0.8, solve for Logb(x)(x^1/3)

  1. Jan 10, 2012 #1
    1. The problem statement, all variables and given/known data

    i got a question on a quiz, but i really did not know how to solve it. the question that was stated was:
    if [itex]\log_b x =0.8[/itex]

    Solve for
    [itex]\log_b x(\sqrt[3]{x})[/itex]


    3. The attempt at a solution

    basically what i attempted to do was
    b0.8 = x

    Logb x + Logb 3√x

    Logb x + 1/3Logb x

    0.8 + 1/3Logb x ?


    EDIT:
    picture attachment
    formatting
     
    Last edited: Jan 10, 2012
  2. jcsd
  3. Jan 10, 2012 #2

    Mark44

    Staff: Mentor

    You're not solving for anything - you are evaluating this expression.

    In any case, is that what you're working with?
    [tex]log_b(x \sqrt[3]{x})[/tex]

    If so, that's different from what you wrote.
     
  4. Jan 10, 2012 #3
    hmm, let me scan it, i dont know how to do all these math symbols
     

    Attached Files:

  5. Jan 10, 2012 #4
    I assume you mean [itex]log[/itex][itex]_{b}[/itex][itex]x[/itex][itex]\sqrt[3]{x}[/itex]?

    Try to get it so you have [itex]klog[/itex][itex]_{b}[/itex][itex]x[/itex], [itex]k[/itex] [itex]constant[/itex]
     
    Last edited: Jan 10, 2012
  6. Jan 10, 2012 #5

    eumyang

    User Avatar
    Homework Helper

    Try to rewrite the expression
    [itex]\log_b(x \sqrt[3]{x})[/itex]
    as a single power, ie.
    [itex]\log_b(x^a)[/itex]
    for some number a. You'll then need the power property of logarithms.
     
  7. Jan 10, 2012 #6
    that would be simple, but its
    [itex]\log_b x( \sqrt[3]{x})[/itex]
     
  8. Jan 10, 2012 #7
    Use the relationship
    [itex]x[/itex][itex]^{\frac{1}{a}}[/itex] = [itex]\sqrt[a]{x}[/itex]

    And [itex]x^{a}[/itex][itex]x^{b}[/itex] = [itex]x^{a+b}[/itex]
     
  9. Jan 10, 2012 #8

    Mentallic

    User Avatar
    Homework Helper

    If it's supposed to be in that form as the question asks, which I suppose should mean

    [tex](\log_bx)(\sqrt[3]{x})[/tex]

    Then simply plug in 0.8 into the first bracket since you're already given that, and solve for x in [itex]\log_bx=0.8[/itex] to find [itex]\sqrt[3]{x}[/itex]
     
  10. Jan 10, 2012 #9
    yea thats my question, how do you solve for x

    when ur given 2 variables?

    b is not given :S

    heres what im doing so far:


    b0.8 = x

    Logb x + Logb 3√x

    Logb x + 1/3Logb x

    0.8 + 1/3Logb x ?
     
    Last edited: Jan 10, 2012
  11. Jan 10, 2012 #10
    That's right. You know [itex]log_{b}x[/itex] = 0.8, so sub that in.
     
  12. Jan 10, 2012 #11
    the only problem is that, the answer should be 1.07 =/
     
  13. Jan 10, 2012 #12

    Mentallic

    User Avatar
    Homework Helper

    Ok so it IS meant to be [tex]log_b(x\sqrt[3]{x})[/tex] as opposed to [tex](log_bx)\cdot \sqrt[3]{x}[/tex]

    Yes, the answer is 1.07, what's the problem? 0.8 + 1/3*0.8 = ?
     
  14. Jan 10, 2012 #13
    ohh, ok i see my mistake now :S

    this is such an easy problem, i was jus over thinking >_<
     
  15. Jan 10, 2012 #14

    Mark44

    Staff: Mentor

    Whoever wrote this problem did not communicate the problem in a clear way. The only way we could understand what was written was to reverse-engineer the solution. The problem should have been written so that it was clear what the argument of the log function was.
     
  16. Jan 10, 2012 #15

    Mentallic

    User Avatar
    Homework Helper

    Yep :wink:

    I was about to ask for the solution if the OP didn't already provide it. This confusion was unnecessarily getting out of hand...

    edit: the answer was written in the attachment. To be honest, it looked like effort to try and understand what was happening on that page so I gave up!
     
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