TITLE: A system of logarithmic equations with parameters
QUESTION [8 upvotes]: $Let$ $a,b,c>0$
$$\begin{cases}\log_a(b^x)=2\\\log_b(c^x)=2\\\log_c(a^x)=5\end{cases}$$
$$x=?$$
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.

REPLY [2 votes]: 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);
$$ \frac{x\cdot\log b}{\log a}\cdot\frac{x\cdot\log c}{\log b}\cdot\frac{x\cdot\log a}{\log c}= 2\cdot 2\cdot 5= 20$$ 
$$ x^3=20,\, x= 20^{1/3}. $$