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Multiple in logarithm question -- find the variable?

  1. Jun 23, 2017 #1
    1. The problem statement, all variables and given/known data
    15_Mat_B_1.1.png

    2. Relevant equations
    $$ \log_{x} a \ means \ 10^x = a $$
    $$ \log_{x} {a^n} = n \log_{x}{a} $$

    3. The attempt at a solution
    $$ \log_{10} {ax} \log_{10}{bx} + 1 = 0 $$
    $$ \ means \log_{10} {ax} = 1 \log_{10}{bx} = -1 \ or \ vice \ versa $$
    $$ \ for \log_{10} {ax} = 1 \ so \ ax \ = 10 \ and \ \log_{10}{bx} = -1 \ so \ bx \ = 0.1 \ or \ vice \ versa $$
    bx/ax = b/a = 0.1/10 = 0.01
    or b/a = 10/0.1 = 100
    so b/a > ???
    how to make range for b/a ? how's the correlation with x?
     
  2. jcsd
  3. Jun 23, 2017 #2

    Mark44

    Staff: Mentor

    No, not true. It just means that ##\log_{10}(ax)\cdot \log_{10}(bx) = -1##. For example, one could be 1/2 and the other could be -2.
    Work from this equation, and use the fact that ##\log(mn) = \log(m) + \log(n)##.
    I haven't worked the problem, but this is what I would do.
     
  4. Jun 23, 2017 #3

    fresh_42

    Staff: Mentor

    What's the formula for ##\log (p \cdot q)##? And what do you get, if you apply it to ##\log(ax)\cdot \log(bx)##
     
  5. Jun 23, 2017 #4

    Ray Vickson

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    No, ##\log_x a## does not mean ##10^x = a##; it means that ##y = \log_x a## is the solution of the equation ##x^y = a##.

    Anyway, your "solution" is all wrong; you are assuming things that need not be true.

    Using the standard result that ##\log_{10}(ax) = \log_{10} a + \log_{10} x##, etc., will help.
     
  6. Jun 23, 2017 #5
    ok, i get
    (loga + log x) . (logb + log x) = -1
    $$ \log_{10}{a} . \log_{10}{b} + 2\log_{10}{x} = -1$$
     
  7. Jun 23, 2017 #6

    Ray Vickson

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    Still wrong. Check your algebra.
     
  8. Jun 23, 2017 #7
    Oh
    .. Loga.logb + (logx)^2 = -1
     
  9. Jun 23, 2017 #8

    fresh_42

    Staff: Mentor

    Let's make it easier and forget about the logarithms for a moment. Say ##a_l = \log a\, , \,x_l = \log x \; , \;b_l=\log b##.
    Now what is ##(a_l + x_l) \cdot (b_l +x_l) + 1 = 0\,##?
     
  10. Jun 23, 2017 #9

    Ray Vickson

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    Are you working things out in detail, or are you just guessing? (Because you still have not gotten it right, although you are getting a bit closer.)
     
  11. Jun 23, 2017 #10
    ab + x(a +b) + x^2 + 1 = 0
    loga . log b + logx(log a + log b) + (logx)^2 + 1 = 0
     
  12. Jun 23, 2017 #11

    fresh_42

    Staff: Mentor

    Yes, and now solve this equation and consider especially the sign of the two solutions and the expression under the root.
     
  13. Jun 26, 2017 #12
    I still don't get it. Can anyone give me hint?
     
  14. Jun 26, 2017 #13

    Dick

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    It's a quadratic equation with ##log(x)## as the variable. Use the quadratic formula!
     
  15. Jul 1, 2017 #14
    let's say I wanna solve the equation ab + x(a +b) + x^2 + 1 = 0
    using abc formula : { -(a+b) +- root [ (a+b)^2 - 4(ab + 1) ] } / 2
    {-a-b +- root [ a^2 + b^2 - 2ab -4 ] } /2
    what am I suppose to do next?
     
  16. Jul 1, 2017 #15
    root1 * root2 = c/a
    root 1 + root 2 = -b / a
    so,
    root 1 * root 2 = c/a = { ab + 1 } / 1 = loga + log b + log 10
    what to do to find root 1 and 2 which is contains log a and log b and contains a and b?
     
  17. Jul 1, 2017 #16

    Ray Vickson

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    You are supposed to figure out what the condition that at least one root is > 0 has to say about a and b.
     
  18. Jul 1, 2017 #17

    Dick

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    If there is going to be a solution then the quantity in the square root needs to be nonnegative. Write down that condition and see if you can simplify it.
     
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