A system of logarithmic equations with parameters $Let$ $a,b,c>0$
$$\begin{cases}\log_a(b^x)=2\\\log_b(c^x)=2\\\log_c(a^x)=5\end{cases}$$
$$x=?$$
So my attempt is just to use the logarithmic definition:
$$\log_a(b^x)=2\iff b^x=a^2$$
By similar logic,
$$a^x=c^5$$
$$c^x=b^2$$
So, if add everything together, we should get:
$$a^x+b^x+c^x=a^2+b^2+c^5$$
Looks to me like x is equal 2 different numbers at the same time which is strange, what am I doing wrong here?
I'm going to be a maths student in the upcoming year, this is taken from the Tel-Aviv university preparation material - shouldn't be too complex.
 A: Using change of base and power formulea, we have 
\begin{eqnarray*}
x \ln b = 2 \ln a \\
x \ln c= 2 \ln b \\
x \ln a = 5 \ln c \\
\end{eqnarray*}
So
\begin{eqnarray*}
x^3 \ln a = 5 x^2 \ln c= 10 x \ln b =20 \ln a  \\
\end{eqnarray*}
Should be easy from here ?
A: $\displaystyle\begin{cases}\log_a(b^x)=2\\log_b(c^x)=2\\log_c(a^x)=5\end{cases}\implies \begin{cases}\frac{\ln(b^x)}{\ln(a)}=2\\\frac{\ln(c^x)}{\ln(b)}=2\\\frac{\ln(a^x)}{\ln(c)}=5\end{cases}$
$\ln(b^x)=2\ln(a)\\\ln(c^x)=2\ln(b)\\\ln(a^x)=5\ln(c)$
adding them all together gives :
$\ln(b^x)+\ln(c^x)+\ln(a^x)=2\ln(a)+5\ln(c)+2\ln(b)$
$\implies x\bigg(\ln(b)+\ln(a)+\ln(c)\bigg)=2\ln(a)+5\ln(c)+2\ln(b) $
$\implies x= \dfrac{2\ln(a)+5\ln(c)+2\ln(b)}{\ln(b)+\ln(a)+ln(c)}$
A: The power of logarithm can be transferred to the front as a coefficient
$$ \log_a b^x=2 \rightarrow \,x\log_a b= \frac{x\cdot\log b}{\log a}=2  $$ irrespective of any chosen base, need even not be mentioned  ( so long as it is real);
$$ \frac{x\cdot\log b}{\log a}\cdot\frac{x\cdot\log c}{\log b}\cdot\frac{x\cdot\log a}{\log c}= 2\cdot 2\cdot 5= 20$$ 
$$ x^3=20,\, x= 20^{1/3}. $$
A: Needless to use the natural logarithm: from
$$\begin{cases}
\log_a b^x=2 \\\log_b c^x=2 \\\log_ca^x=5
\end{cases} \iff
\begin{cases}
x\log_a b=2 \\x\log_b c=2 \\x\log_ca=5
\end{cases} $$
you deduce readily that
$$x^3(\underbrace{\log_ab\,\log_b c\,\log_c a}_{=1})=20.$$
