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

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Homework Help Overview

The discussion revolves around proving the equality of a complex expression involving logarithmic relationships among variables x, y, and z, given specific conditions related to parameters a, b, and c.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various methods to manipulate logarithmic expressions and question the relationships between the variables. Some participants suggest working backwards from the desired equality, while others express confusion about the introduction of the constant lambda and its implications.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and seeking clarification on specific steps. Some have provided hints and guidance, while others are still grappling with the concepts involved.

Contextual Notes

There is a noted difference in familiarity with logarithmic concepts among participants, with some expressing that they have not yet studied logs in their curriculum.

Govind_Balaji
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Homework Statement

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

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 m= b-c, n= c-athen a-b=-(m+n)

By cross multiplication, I got

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

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

I am struggling to substitute here.

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

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

I am struggling then after. Can anyone give me a hint.
 
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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?
 
Pranav-Arora said:
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 b+c-a=0 \text{then } c+a-b=0 \text{ and } a+b-c=0
 
Govind_Balaji said:
Are you saying I have to prove b+c-a=0 \text{then } c+a-b=0 \text{ and } a+b-c=0

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##?
 
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Pranav-Arora said:
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?
 
Govind_Balaji said:
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? :-p

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:
 
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Yes, now I will try.
 
Pranav-Arora said:
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 can't 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.
 
Govind_Balaji said:
##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##)
 
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  • #10
I get

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

Am I right till now?
 
  • #11
Govind_Balaji said:
I get

<br /> <br /> \\\log x^{b+c-a}=\lambda\frac{b+c-a}{b-c}\\\\<br /> \\\log y^{c+a-b}=\lambda\frac{c+a-b}{c-a}\\\\<br /> \\\log z^{a+b-c}=\lambda\frac{a+b-c}{a-b}\\\\<br />
Are you sure that they should be in denominator?
 
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  • #12
Pranav-Arora said:
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
Govind_Balaji said:
I am just 9th grade.
You are at 9th grade and you have studied logs?
I haven't still studied it. :cry:
 
  • #14
adjacent said:
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?:-p
 
  • #16
Govind_Balaji said:
I saw in your profile. You have done PhD and you have said that you are a professor.

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

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