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?

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

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?
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 %26lt;%26gt; 0


a=1


a=-1, xyz is odd.
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Reply: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).
Reply: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.
Reply: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!!
Reply:a^x=b


[a^x]^y=b^y=c


[[a^x]^y]^z=c^z=a


a^xyz=a


xyz=1 QED.
Reply: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
Reply:a^x = b -%26gt;


a^x^y = b^y = c -%26gt;


a^x^y^z = c^z = a -%26gt;


a^(xyz) = a^1 -%26gt;


xyz = 1