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Prove a complex expression = 1, if log of the complex terms are equal

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data


    If [itex]\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b},[/itex] prove that [itex]x^{b+c-a}.y^{c+a-b}.z^{a+b-c}=1[/itex]


    2. Relevant equations



    3. The attempt at a solution

    I solved a question similar to it in a way. So I tried this in the same way. The only difference between the two questions is that except a,b and c, the denominators are also x, y and z.
    I didn't know if it would work.

    So I tried like this.
    Let [itex] m= b-c, n= c-a [/itex]then [itex] a-b=-(m+n)[/itex]

    By cross multiplication, I got

    [itex]z^m=x^{-(m+n)}\\z^n=y^{-(m+n)}\\x^n=y^m\\\\\Rightarrow z=1/xy, x=1/yz, y=1/xz[/itex]

    In my old question it was to prove [itex]x^x.y^y.z^z=1[/itex]. So I substituted easily.

    I am struggling to substitute here.

    So I tried writing [itex]x^{b+c-a}.y^{c+a-b}.z^{a+b-c}[/itex] it easier by using m and n.

    [itex]x^{b+c-a}.y^{c+a-b}.z^{a+b-c}=x^{2n+m+a}.y^{a-m}.z^{a+m}[/itex]

    I am struggling then after. Can anyone give me a hint.
     
    Last edited: May 20, 2014
  2. jcsd
  3. May 20, 2014 #2
    Try working backwards.

    You need to prove ##x^{b+c-a}.y^{c+a-b}.z^{a+b-c}=1##, this is equivalent to showing ##\log x^{b+c-a}+\log y^{c+a-b}+\log z^{a+b-c}=0##, do you see why? If so, can you figure out the individual terms and show that their sum is zero?
     
  4. May 20, 2014 #3

    Are you saying I have to prove [itex]b+c-a=0 \text{then } c+a-b=0 \text{ and } a+b-c=0
     
  5. May 20, 2014 #4
    No, I didn't mean that. Equate the equality to ##\lambda##. Can you find ##\log x^{b+c-a}## in terms of ##\lambda, a, b## and ##c##?
     
  6. May 20, 2014 #5
    I am just 9th grade. I don't know anything about lambda other than it is a symbol for wavelength of light. I don't know the value of lambda. Can you explain clearly?
     
  7. May 20, 2014 #6
    Wavelength? :tongue:

    I meant ##\lambda## as some constant, you can even name it ##\alpha## if you wish to. What I meant was this:
    $$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=\lambda$$
    Can you make sense of this now? :smile:
     
  8. May 20, 2014 #7
    Yes, now I will try.
     
  9. May 21, 2014 #8
    I cant understand how to prove this. expressing ##\log x^{b+c-a}## in terms of ##\lambda, a, b## and ##c##, ##10^{\lambda}=x^{b+c-a}##.

    Or I tried as I said in the first post of this thread.

    Can you give a small hint if you have already solved it.
     
  10. May 21, 2014 #9
    I don't see how you get that.

    You have the following:
    $$\frac{\log x}{b-c}=\lambda$$
    Can you find ##\log x^{b+c-a}## from here? (Hint: Multiply both the sides by ##b+c-a##)
     
  11. May 21, 2014 #10
    I get

    [itex]

    \\\log x^{b+c-a}=\lambda\frac{b+c-a}{b-c}\\\\
    \\\log y^{c+a-b}=\lambda\frac{c+a-b}{c-a}\\\\
    \\\log z^{a+b-c}=\lambda\frac{a+b-c}{a-b}\\\\

    \\\log\left(x^{b+c-a}. y^{c+a-b}.z^{b+c-a}\right )=\lambda\frac{b+c-a}{b-c}+\lambda\frac{c+a-b}{c-a}+\lambda\frac{a+b-c}{a-b}

    [/itex]

    Am I right till now?
     
  12. May 21, 2014 #11
    Are you sure that they should be in denominator?
     
  13. May 21, 2014 #12
    Thank you, a small careless mistake.

    The denominators should be in numerators.

    I add them. I got 0.

    Removing the log, I will get them equal to 0.

    Thank you very much.
     
  14. May 21, 2014 #13

    adjacent

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    Gold Member

    You are at 9th grade and you have studied logs?
    I haven't still studied it. :cry:
     
  15. May 21, 2014 #14
    I saw in your profile. You have done PhD and you have said that you are a professor.

    Why are you kidding me?:tongue2:
     
  16. May 21, 2014 #15

    DrClaude

    User Avatar

    Staff: Mentor

  17. May 21, 2014 #16

    adjacent

    User Avatar
    Gold Member

    You can see my age(15 years). I am in Gr.10 :smile:
     
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