Given $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ show that $x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$

Question

Given:

$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}$$

We have to show that :

$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$

I made three equations using cross multiplication :

$$1.~~x^{c-a}=y^{b-c}$$ $$2.~~y^{a-b}=z^{c-a}$$ $$3.~~z^{b-c}=x^{a-b}$$

How do I proceed hereafter? If I multiply the equations, one variable goes away from exponents.

Thank you.

A couple of answers posted below assume the base of the logarithm is $e$, and that is what is normally presumed when the base is not specified. However, in this case it doesn't matter what the base is, as long as all three logarithms are to the same base. I've posted an answer below that shows this. ${}\qquad{}$ — Michael Hardy, Apr 26 '15 at 14:24

score 13 · Accepted Answer · answered Apr 26 '15 at 14:11

13

We have $$\dfrac{\log(x)}{b-c} = \dfrac{\log(y)}{c-a} = \dfrac{\log(z)}{a-b} = t$$ This gives us $$x=e^{t(b-c)}, y = e^{t(c-a)} \text{ and }z = e^{t(a-b)}$$ Hence, \begin{align} x^{b+c-a}\cdot y^{c+a-b} \cdot z^{a+b-c} & = e^{t\left((b-c)(b+c-a) + (c-a)(c+a-b) + (a-b)(a+b-c)\right)}\\ & = e^{t(b^2-c^2-ab+ac + c^2 - a^2 -bc + ba + a^2 - b^2 - ac + bc)} = e^0 = 1 \end{align}

answered Apr 26 '15 at 14:11

Adhvaitha

20,259

This is nice! very short method! but can i proceed in way i started? can it be done that way? Thanks! – Max Payne Apr 26 '15 at 14:17

score 10 · Answer 2 · answered Apr 26 '15 at 15:28

10

If you want to use your equations, here is a method.

Multiplying the equations together, we obtain: $$x^{c-a}y^{a-b}z^{b-c}=y^{b-c}z^{c-a}x^{a-b}$$ which gives after reordering: $$x^{b+c-a}y^{c+a-b}z^{a+b-c}=x^a y^b z^c$$ Therefore it suffices to show that $x^a y^b z^c = 1$.

Your first and third equations give $y = x^{\frac {c-a}{b-c}}, z = x^{\frac{a-b}{b-c}}$. This gives us: $$x^a y^b z^c = x^a x^{\frac {c-a}{b-c}\times b} x^{\frac{a-b}{b-c}\times c} = x^{a + \frac{bc-ba+ca-bc}{b-c}} = x^{a-a} = x^0 = 1 $$

QED.

answered Apr 26 '15 at 15:28

yoann

1,363

1

Excellent. From this it also follows that $(xyz)^{a+b+c}=1$ and therefore $xyz = 1$ or $a+b+c=0$. – M. Wind Apr 26 '15 at 15:44
Really nice! I wa thinking that it wasn't possible! +(1). Thank you very much! – Max Payne Apr 27 '15 at 14:02

score 5 · Answer 3 · answered Apr 26 '15 at 14:12

5

Given: $$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}=\lambda$$ we have: $$ x = e^{\lambda(b-c)},\quad y=e^{\lambda(c-a)},\quad z=e^{\lambda(a-b)}, $$ hence: $$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = \exp\left(\lambda\cdot\sum_{cyc}\left(b^2-c^2-a(b-c)\right)\right)=\exp(0)=1.$$

answered Apr 26 '15 at 14:12

Jack D'Aurizio

353,855

It is interesting that when I wrote the answer, I was debating between using $t$ or $\lambda$ and decided in favor of $t$ to avoid additional typesetting. The interesting part is that how certain notations people always use, like for instance $\lambda$ in this case. – Adhvaitha Apr 26 '15 at 14:14

score 4 · Answer 4 · answered Apr 26 '15 at 14:23

4

Hint

If you have to show that $$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$ taking logarithms of both sides means that you have to show that $$(b+c-a)\log(x)+(c+a-b)\log(y)+(a+b-c)\log(z)=0$$ Now, use what user17762 and Jack D'Aurizio answered.

answered Apr 26 '15 at 14:23

Claude Leibovici

260,315

Thank you i think it is much simpler way. but i wanted to know if i could proceed using the three equations that i derived.. – Max Payne Apr 26 '15 at 14:28

score 2 · Answer 5 · answered Apr 26 '15 at 14:22

2

$\dfrac{\log x}{b-c}$ is equal to $\log\left(x^{1/(b-c)}\right)$ regardless of what the base of the logarithm is. Hence we have $$ \log\left(x^{1/(b-c)}\right) = \log\left(y^{1/(c-a)}\right) = \log\left(z^{1/(a-b)}\right). $$ Since the logarithm function is one-to-one, this entails $$ x^{1/(b-c)} = y^{1/(c-a)} = z^{1/(a-b)}. $$ Raising both sides of $x^{1/(b-c)} = y^{1/(c-a)}$ to the power $(b-c)(c-a)$ yields $$ x^{c-a} = y^{b-c} $$ and the other two equalities are derived similarly.

answered Apr 26 '15 at 14:22

Michael Hardy

1

Yes I already derived that equalities. I wanted to proceed from there on .. – Max Payne Apr 26 '15 at 14:26
Sorry --- I stopped too early; maybe I'll be back in an hour or so to add something. – Michael Hardy Apr 26 '15 at 14:53

FranMowinckel · Answer 6 · 2015-04-26T16:47:14.087

First let's assume that there are no indeterminations:

$x > 0 \land y > 0 \land z > 0 \land a \neq b \neq c$

And we have to prove that $k = 1$ in:

$x^{b+c−a}⋅y^{c+a−b}⋅z^{a+b−c} = k$

Using the asker equations from cross multiplication:

$x^{b+c−a}⋅x^{a-c}⋅y^{a}⋅z^{a}⋅x^{a-b} = x^a⋅y^a⋅z^a = (xyz)^a = k$

From this method, we can also obtain:

$(xyz)^b = k$

and

$(xyz)^c = k$

But then:

$k^{\frac{1}a} = k^{\frac{1}b} = k^{\frac{1}c} \implies k = 1$

What's more:

$xyz = 1$

score 0 · Answer 7 · edited Jun 12 '20 at 10:38

0

The product of the three equalities you have given is:
$x^{c-a} y^{a-b} z^{b-c} = y^{b-c} z^{c-a} x^{a-b}$

Grouping similar variables together,

$(x^{c-a} x^{b-a}) (y^{a-b} y^{c-b}) (z^{b-c} z^{a-c}) = 1$
$x^{-2a+b+c} \times y^{a-2b+c} \times z^{a+b-2c} = 1$

edited Jun 12 '20 at 10:38

Community

1

answered Apr 26 '15 at 14:43

yellow_submarine

568

Thanks but it isnt correct answer! Look at LHS of your final statement, and the equation to be proved. There is a coefficient of -2 where it should be -1 – Max Payne Apr 26 '15 at 14:46
Have you considered the option that the proposed answer is not right? – yellow_submarine Apr 26 '15 at 14:47
1

Actually, you have demonstrated that another relation between $(x,y,z)$ and $(a,b,c)$ must be valid: $x^{-a}y^{-b}z^{-c} = 1$. – M. Wind Apr 26 '15 at 15:25
Exactly! the question asked something else. – Max Payne Apr 27 '15 at 13:15

Given $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ show that $x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$

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