$(ab)^2=(bc)^4=(ca)^x=abc$ Then what is the value of $x$? 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?
 A: 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 ;-)
A: 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 $x=\frac12$
A: 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 $x=4$.
A: $2(\log a+\log b) = 4(\log b+\log c) = x(\log c+\log a ) = \log a + \log b+\log c$
Separating the equalities into two,
$$
2(\log a+\log b)= \log a+\log b+\log c \Rightarrow \log a + \log b - \log c=0 \quad(i)
$$
$$
4(\log b+\log c)=\log a+\log b+\log c\Rightarrow 3(\log b+\log c)-\log a=0 \quad(ii)
$$
From equation (i) and (ii), upon solving, we get,
$$
\log a=(3/2)\log c,~\log b=-(\log c)/2 \quad(iii)
$$
$$
4(\log b+\log c)=x(\log c+\log a) \quad(iv)
$$
By putting (iii) into (iv),
$$
-2\log c+4\log c=x\log c+(3x/2)\log c,
\\
2\log c=(5x/2)\log c.
$$
By dividing $\log c$ on both sides,
$$
2=(5x/2)\Rightarrow x=4/5.
$$
Thanks.
