Question

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Answer 1

\[\huge \frac{ logx }{ b-c } = \frac{ logy }{ c-a } = \frac{ logz }{ a-b }\]
Show That
\[\huge x ^{a}y ^{b}z ^{c}=1\]

Answer 2

I proved xyz=1

Answer 3

From the equality given, we can get 3 equations.
Take 2 equations at once.
These are the equations when we cross multiply; we get:

c.log x - a.log x = b.log y - c.log y

From this equation we can get:

log (x^c / x^a) = log (y^b / y^c)

x^(c-a) = y^(b-c)
Or

x^(c-a)/y^(b-c) = 1
You should do the same with the other equations too:

a.log y- b.log y = c.log z - a.log z
a.log x - b.log x = b.log z - c.log z

And then you'd have to multiply all the equations. And I THINK it should give you the answer. Just try it once.

Did you get what I did there? ^ :)

Answer 4

log (y^a /y^b) = log ( z^c/ z^a)
y^(a-b)/z^(c-a) = 1

Answer 5

Yes! Excellent! Now you've got to do the same with the third equation.
A then multiply the new equations.
But did you understand how the given equations are turning to the equations at need?
That is more important.

Answer 6

Yeah i know the log properties but i didn't get that intuition
BTW are you in CBSE board or SSE
It seems you are in 12 th

Answer 7

Yes. I am in 12th in a CBSE board.

Answer 8

a.log x - b.log x = b.log z - c.log z
logx^a - logx^b = logz^b - logz^c
x^(a-b) = z^(b-c)

Answer 9

Yes, now divide till you get the RHS as 1.
And then, multiply all the equations you get.

Hopefully, you wouldn't get 1=1. :P

Answer 10

You wouldn't!
NOTE: As it turns out, there are 3 equations with all the variables. And thus, one a,b,c is bound to be left out. :D

Answer 11

\[\huge \frac{ x ^{a-b} }{ z ^{b-c} }=\frac{ y ^{a-b} }{ z ^{c-a} }=\frac{ x ^{c-a} }{ y ^{b-c} }\]

Answer 12

After that

Answer 13

No quite like that.

I was telling you take the other individual equations [The ones which have a '1' on the RHS]
And then multiply all of them because, then the RHS will finally have '1'.

Answer 14

I am not getting you sorry

Answer 15

Which equations you are talking about

Answer 16

Here. Like I showed you, I had got
\[\frac{x^{c-a}}{y^{b-c}} = 1\]
Then you found this: \[\frac{y^{a-b}}{z^{c-a}} = 1\]
And then again from the 3rd equation, you got this: \[\frac{x^{a-b}}{z^{b-c}} = 1\]

Now just multiply these 3 equations, and you'd have your required answer.
But you should check @ranga 's solution too, as he may have a shorter one. :)

Answer 17

\[\Large \frac{ logx }{ b-c } = \frac{ logy }{ c-a } = \frac{ logz }{ a-b } = k\]
log(x) = k(b-c); alog(x) = ka(b-c)
log(y) = k(c-a); blog(y) = kb(c-a)
log(z) = k(a-b); clog(z) = kc(a-b)

alog(x) + blog(y) + clog(z) = k(ab - ac + bc - ba + ca - cb) = 0
log(x^a * y^b * z^c) = log(1)
x^a * y^b * z^c = 1

Answer 18

Arrghh! I always forget to use the constant to do these. -__-
Use that! ^

Answer 19

Interesting! Thank you all of you

Answer 20

You'd get the answer, either ways. :)

Answer 21

You are welcome.

Answer 22

@AkashdeepDeb Did u give IIT jee (mains) this year?

Answer 23

I did study for it. But no. I'd give it next year.
I just started 12th. :P