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

  • #1

Homework Statement




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]


Homework Equations





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:

Answers and Replies

  • #2
3,816
92
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?
 
  • #3
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?


Are you saying I have to prove [itex]b+c-a=0 \text{then } c+a-b=0 \text{ and } a+b-c=0
 
  • #4
3,816
92
Are you saying I have to prove [itex]b+c-a=0 \text{then } c+a-b=0 \text{ and } a+b-c=0[/itex]

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##?
 
  • #5
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##?

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?
 
  • #6
3,816
92
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?

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
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##?

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.
 
  • #9
3,816
92
##10^{\lambda}=x^{b+c-a}##.

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##)
 
  • #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?
 
  • #11
3,816
92
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}\\\\
[/itex]
Are you sure that they should be in denominator?
 
  • #12
Are you sure that they should be in denominator?

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.
 
  • #13
adjacent
Gold Member
1,553
63
I am just 9th grade.
You are at 9th grade and you have studied logs?
I haven't still studied it. :cry:
 
  • #14
You are at 9th grade and you have studied logs?
I haven't still studied it. :cry:

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
adjacent
Gold Member
1,553
63
I saw in your profile. You have done PhD and you have said that you are a professor.

Why are you kidding me?:tongue2:
You can see my age(15 years). I am in Gr.10 :smile:
 

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