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Another question about log

  1. Nov 2, 2004 #1
    Say you have:
    [itex]\frac{\log(x)}{r\log{x}} = y[/itex]

    r and y can be any given number.

    Is there any way to solve this for x?


    [itex]\frac{\log(x)}{1.6\log{x}} = 20[/itex]

    I'm getting this because I'm trying to calculate [itex]f(x) = \frac{x}{a}[/itex] where f(x) is of the form [itex]f(x) = cx^r[/itex]
    [itex]a > 1[/itex]
    Last edited: Nov 2, 2004
  2. jcsd
  3. Nov 2, 2004 #2
    I don't get this, sorry. correct me if i am wrong here but i have :

    [tex]\frac {\log(x)}{r\log(x)} = y[/tex] but doesn't the left hand side yield
    [tex]\frac {1}{r}[/tex] ???

    i am missing the point here...you sure about this equation ????

  4. Nov 2, 2004 #3
    Well, I'm trying to find x for [itex]f(x) = \frac{x}{a}[/itex]

    where f is of the form [itex]cx^r[/itex]

    I tried setting [itex]x = cx^r[/itex] and taking the log to find x.

    I was wondering if it was possible to find x that way, or maybe if there is another way?
  5. Nov 2, 2004 #4
    [itex]cx^r = \frac {x}{a}[/itex]
    [itex]cx^r - \frac {x}{a} = 0 [/itex]
    [itex]x * (cx^{r-1} - \frac {1}{a}) = 0 [/itex]
    [itex]cx^{r-1} = \frac {1}{a}[/itex] if x is not 0
    [itex]x^{r-1} = \frac {1}{ca}[/itex] if x is not 0
    [itex]x = \sqrt [r - 1] {\frac {1}{ca}}[/itex] if x is not 0

  6. Nov 2, 2004 #5
    what do you say about that ???

  7. Nov 2, 2004 #6
    Hm, there must be a more elegant way?
  8. Nov 2, 2004 #7
    trust me , this is the most easy way out. working with log will make you go around in circles

  9. Nov 2, 2004 #8
    Hehe, ok, works for me.

    I'm finding myself confused looking for the more complex ways to solve stuff, when regular math will do the job. Oh well :)

    thanks a lot.
  10. Nov 3, 2004 #9

    matt grime

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    Science Advisor
    Homework Helper

    If you did your logs properly it would also drop out:


    then, assuming x and every thing else for that matter, is positive,

    logc + rlogx = logx - loga

    (r-1)logx= -loga - log c

    log x = -log(ac) * 1/(r-1)

    or x = (ac)^{1/(1-r)

    exactly as marlon showed without needing to use logs.
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