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.

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 $$x^3(\underbrace{\log_ab\,\log_b c\,\log_c a}_{=1})=20.$$

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 ?