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Given that $(ab)^2=(bc)^4=(ca)^x=abc$ Then what is the value of $x$?

$2(\log a+\log b)=4(\log b+\log c)=x(\log c+\log a)=\log a+\log b+\log c$

Then I am lost, any other easier way to solve?

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$a,b,c$ are positive reals and $abc\ne 1$ –  miosaki Nov 28 '13 at 13:27

3 Answers 3

up vote 2 down vote accepted

Taking logarithm gives: $$2(\log a+\log b)=4(\log b+\log c)=x(\log c+\log a)=\log a+\log b+\log c$$

then taking $2(\log a+\log b)=\log a+\log b+\log c$

and $4(\log b+\log c)=\log a+\log b+\log c$

we would get $\log a+\log b-\log c=0$ & $3\log b+3\log c-\log a=0$ and by solving these two equations we get $\log b=-\log c$

similarly $\log a=-\log b$ then the solution becomes obvious......as $x=\frac12$

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There is a mistake after "by solving these equations...": we get $2\log b=-log c$ and $\log a=-3\log b$. This also changes the value of x –  Frédéric Grosshans Oct 23 at 16:30

Hint: Split these two equalities into

$$(ab)^2 = (ca)^x$$ and $$(bc)^4 = (ca)^x$$

Then use $\log$ on both equations and see what happens ;-)

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Let’s try to solve this without logarithms, as simply as possible. This allows to extend the solution to negative and complex values of $a$,$b$ and $c$, and includes a more interesting set of solution in the specific case of $|b|=1$.

If $abc=0$, then at least two elements of $\{a,b,c\}$ are $0$, and $x$ can take any value, except $0$ if $0^0=1$.

From now on $abc\neq 0$ : $$(ab)^2=abc \Rightarrow c=ab$$ replacing $c$ by its value transforms the equalities into : $$(ab)^2=\left(ab^2\right)^4=\left(a^2b\right)^x$$ The first equality then gives $$a^2b^6=1$$ Replacing $a^2$ by $\frac1{b^6}$, we have then $$\frac1{b^4}=\left(\frac1{b^5}\right)^x \Leftrightarrow b^4=b^{5x}.$$

If $b=1$, this is true for all $x$ and $a=c=\pm1$

Otherwise, if $|b|≠1$, one has $\boxed{x=\frac45}$ and $a=\pm\frac1{b^3}$, $c=\pm\frac1{b^2}$. This is the solution you were looking for.

But, for $|b|=1$ and $b≠1$, the solution above is still true, but it is not the only one. Let’s define $β≠0$ by $b=e^{iβ}$. The conditions on $x$ then becomes $$5xβ=4β+2kπ, k∈\mathbb Z$$ giving the set of solutions $$x=\frac45+\frac{2kπ}{5β}, k∈\mathbb Z.$$

This includes, for example the non trivial solution where $a=b=i$, $c=-1$ and $x=4$.

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