A system of logarithmic equations with parameters

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 ?

$\displaystyle\begin{cases}\log_a(b^x)=2\\log_b(c^x)=2\\log_c(a^x)=5\end{cases}\implies \begin{cases}\frac{\ln(b^x)}{\ln(a)}=2\\\frac{\ln(c^x)}{\ln(b)}=2\\\frac{\ln(a^x)}{\ln(c)}=5\end{cases}$

$\ln(b^x)=2\ln(a)\\\ln(c^x)=2\ln(b)\\\ln(a^x)=5\ln(c)$

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) $

$\implies x= \dfrac{2\ln(a)+5\ln(c)+2\ln(b)}{\ln(b)+\ln(a)+ln(c)}$