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

## Homework Statement

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

## 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 $m= b-c, n= c-a$then $a-b=-(m+n)$

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.

Last edited:

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?

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 $b+c-a=0 \text{then } c+a-b=0 \text{ and } a+b-c=0 Are you saying I have to prove [itex]b+c-a=0 \text{then } c+a-b=0 \text{ and } a+b-c=0$

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##?

1 person
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?

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?

1 person
Yes, now I will try.

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.

##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##)

1 person
I get

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

Am I right till now?

I get

$\\\log x^{b+c-a}=\lambda\frac{b+c-a}{b-c}\\\\ \\\log y^{c+a-b}=\lambda\frac{c+a-b}{c-a}\\\\ \\\log z^{a+b-c}=\lambda\frac{a+b-c}{a-b}\\\\$
Are you sure that they should be in denominator?

1 person
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.

Gold Member
You are at 9th grade and you have studied logs?
I haven't still studied it.

You are at 9th grade and you have studied logs?
I haven't still studied it.

I saw in your profile. You have done PhD and you have said that you are a professor.

Why are you kidding me?:tongue2: