Can anybody help me out with this math problem? I got to prove that if a^x=b, b^y=c, and c^z=a, then xyz =1?

Question

My teacher told me that it is a simple problem and that I dont need to use logarithm for this, so please dont give me answers by solving this problem using log, my teacher would skin me alive!
P.S: But of course, she probably would skin me alive if she knew what I'm doing to get the answer! :P

Answer 1

Step by Step

plug a^x=b into b^y=c, and get

(a^x)^y=c

a^(xy)=c

plug into c^z=a, and get

a^(xyz)=a

Now, there are 3 solutions other than xyz=1:

a=0, and xyz <> 0
a=1
a=-1, xyz is odd.

Answer 2

Substitute each variable on the right-hand sides into the left-hand side of the next equation.

a^x = b and b^y = c therefore (a^x)^y = c.

(a^x)^y = c and c^z = a therefore ((a^x)^y)^z = a.

Simplify using the rule about exponents.

a^(xyz) = a. Definitely true when xyz = 1.

However, the last equation is also true when a = 0 and xyz = anything you want, so it was not a very good question.

Answer 3

Well,your teacher is right .It looks easy !The solution is straight forward :

1 ) b^y=c and a^x=b ,therefore you arrive at (a^x)^y = c

2 ) c^z = a and (a^x)^y = c ,therefore you arrive at a^xyz = a

So xyz = 1

Hope it helps!!

Answer 4

Take logs of both sides of each equation
xlna = ln b
ylnb = ln c
zlnc = ln a

multiply all 3 equations

xyz(lna)(lnb)(lnc) = (lnb)(lnc)(lna)
xyz = 1

Edit: Sorry didn't see the log comment. Still the restriction is odd since using exponents instead of logs is akin to using multiplication instead of addition (i.e. that a*n is a multiplied by n and not a added to itself n times).

Answer 5

a^x = b ->
a^x^y = b^y = c ->
a^x^y^z = c^z = a ->
a^(xyz) = a^1 ->
xyz = 1

Answer 6

a^x=b
[a^x]^y=b^y=c
[[a^x]^y]^z=c^z=a
a^xyz=a
xyz=1 QED.

Answer 7

if a^x=b
b^y=c, and c^z=a

(c^z)^x = b
(b^y)(^z)^x = b
b^(xyz = b
yzx = 1