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If $$a=b^c, \quad b=c^a, \quad \text{and} \quad c=a^b$$ prove that $$abc=1.$$

My Attempt;

Given, $$a=b^c$$ $$b=c^a$$ $$c=a^b$$

Now,

$$ \begin{align} \text{L.H.S.} & =abc \\ & = b^c\cdot b\cdot a^b \\ & =b^c\cdot b\cdot b^{bc} \\ & =b^{c+1+bc} \\ & \,\,\,\vdots \end{align} $$

I got struck from here.

Please help to complete.

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  • $\begingroup$ $b^{cba}=a^{ba}=c^a=b$ $\endgroup$
    – Akiva Weinberger
    Mar 23, 2016 at 2:39

5 Answers 5

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In $a=b^c$, substitute $b=c^a$, and you get $a=(c^a)^c=c^{ac}$. Next substitute $c=a^b$ and get $a=(a^b)^{ac}=a^{abc}$. So $a^{1}=a^{abc}$. Either $abc=1$ or $a=1$.

If $a=1$ then $c=a^b=1$ and $b=c^a=1$, so $abc=1$ again.

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  • $\begingroup$ How do you know $a>0$? $a=b=c=-1$ works $\endgroup$
    – user223391
    Mar 23, 2016 at 0:32
  • $\begingroup$ @ZacharySelk He didn't know. He said "If", which it indeed works. $\endgroup$
    – Simply Beautiful Art
    Mar 23, 2016 at 0:40
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$$abc = a^bb^cc^a$$ $$\implies \log(abc) = \log(a^bb^cc^a)$$ $$\implies \log(abc) = \log b^c + \log c^a + \log a^b$$ $$\implies \log(abc) = c\log b + a\log c + b\log a$$ $$\implies \log(abc) = c\log c^a + a\log a^b + b\log b^c$$ $$\implies \log(abc) = ac\log c + ab\log a + bc\log b$$ $$\implies \log(abc) = ac\log a^b + ab\log b^c + bc\log c^a$$ $$\implies \log(abc) = abc\log a + abc\log b + abc\log c$$ $$\implies \log(abc) = abc(\log a + \log b + \log c)$$ $$\implies \log(abc) = abc\log(abc)$$ $$\implies 1 = abc$$

Not the shortest proof, but I consider it by far the most beautiful I could think of due to the symmetry in the logarithm steps!

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  • $\begingroup$ How do you know $a>0$? $a=b=c=-1$ works $\endgroup$
    – user223391
    Mar 23, 2016 at 0:31
  • $\begingroup$ @ZacharySelk you are absolutely right, and I saw the post by SimpleArt prior to answering. The OP asks for a proof that it equals 1, implying that $abc = 1$ must be true. I implicitly assume this means that $a,b,c > 0$, though you are correct - The OP never says so explicitly. However, if I don't accept that each variable is non-negative then the OP's conclusion cannot be reached! $\endgroup$
    – Brevan Ellefsen
    Mar 23, 2016 at 0:34
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Another answer that involves logarithms.

if $a=b^c$, $b=c^a$, $c=a^b$ then we may write that

$c=\log_{b}(a)$,$a=\log_{c}(b)$ and $b=\log_{a}(c)$

Hence we have that $abc=\log_{b}(a)\log_{c}(b)\log_{a}(c)$. Then by using change of base rule we get

$abc=\frac{\log_{e}(a)}{\log_{e}(b)}\frac{\log_{e}(b)}{\log_{e}(c)}\frac{\log_{e}(c)}{\log_{e}(a)}$

Notice that all the logarithms on the right hand side cancel out and we get the desired result.

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$$ a^1 = a = b^c = (c^a)^c = c^{ac} = (a^b)^{ac} = a^{bac}, $$ so $$ a^1 = a^{bca}. $$ Now take base-$a$ logarithms of both sides.

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  • $\begingroup$ How do you know $a$ is positive? Couldn't $a=b=c=-1$ work? $\endgroup$
    – user223391
    Mar 23, 2016 at 0:30
  • $\begingroup$ @ZacharySelk You don't actually know, the solution $a=b=c=-1$ is simply due to the way you take the log of both sides, which has problems when we allow multiple branches of the log. $\endgroup$
    – Simply Beautiful Art
    Mar 23, 2016 at 0:38
  • $\begingroup$ @ZacharySelk : Generally I assume a number raised to a real power (which may not be an integer power) is nonnegative. Raising a negative number to a non-integer power can be problematic. $\qquad$ $\endgroup$
    – Michael Hardy
    Mar 23, 2016 at 5:30
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Actually, I note that $a=b=c=-1$ works just as well, which contradicts everyone's statements that $abc=1$.

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  • $\begingroup$ You are absolutely right, and I saw your post prior to answering. The OP asks for a proof that it equals 1, implying that $abc = 1$ must be true. I implicitly assume this means that $a,b,c > 0$, though you are correct - The OP never says so explicitly. However, if I don't accept that each variable is non-negative then the OP's conclusion cannot be reached, so I did so anyway! Your observation is critical though - the OP must add some criterion on the domain of $a$, $b$, and $c$ $\endgroup$
    – Brevan Ellefsen
    Mar 23, 2016 at 0:35
  • $\begingroup$ @BrevanEllefsen Yeah, I saw your answer as well, and it most certainly was neat and well made. I just happened upon this answer by chance. ;D Noting that $|a|=|b|=|c|=1$, but with the restriction of the solutions being positive, of course, everyone else is just as right. :) $\endgroup$
    – Simply Beautiful Art
    Mar 23, 2016 at 0:38

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