If $x=\log_{2a} a$, $y=\log_{3a} 2a$, and $\log_{4a} 3a$, prove that: If $x=\log_{2a} a$, $y=\log_{3a} 2a$, and $\log_{4a} 3a$, prove that:
$$xyz+1=2yz$$
I can't get any idea. please help.
Thanks! 
 A: Note the following formula: $$\log_{\alpha}(\beta) = \frac{\log \beta}{\log \alpha}$$
Thus, we get, $$xyz = \frac{\log a}{\log 4a}\, \, \text{and} \,\, yz = \frac{\log 2a}{\log 4a}$$
We thus get: $$xyz + 1 = \frac{\log a + \log 4a}{\log 4a}=\frac{\log 4a^2}{\log 4a}=\frac{\log(2a)^2}{\log 4a}=\frac{2\log 2a}{\log 4a}=2yz$$
and we are done!
A: For $a>0$, $a\neq\frac{1}{2}$, $a\neq\frac{1}{3}$ and $a\neq\frac{1}{4}$ we obtain: $$2yz-xyz=\frac{2\ln2a\ln3a}{\ln3a\ln4a}-\frac{\ln{a}\ln2a\ln3a}{\ln2a\ln3a\ln4a}=$$
$$=\frac{2\ln2a}{\ln4a}-\frac{\ln{a}}{\ln4a}=\frac{\ln\frac{4a^2}{a}}{\ln4a}=1.$$
A: There’s probably a faster way but this will work: 
$$xyz+1 = \frac{\log(a)}{\log(2a)} \frac{\log(2a)}{\log(3a)} \frac{\log(3a)}{\log(4a)} + 1 = \frac{\log(a)}{\log(4a)} + 1 $$
and 
$$2yz = 2\frac{\log(2a)}{\log(3a)} \frac{\log(3a)}{\log(4a)} = 2 \frac{\log(2a)}{\log(4a)} = \frac{\log(4a^2)}{\log(4a)} = \frac{\log(a) + \log(4a)}{\log(4a)} = \frac{\log(a)}{\log(4a)} + 1 = xyz+1.$$
A: \begin{eqnarray*}
x= \log_{2a}(a) =\frac{ \ln a}{\ln2 + \ln a} \\
y= \log_{3a}(2a) =\frac{\ln 2 + \ln a}{\ln3 + \ln a} \\
z= \log_{4a}(3a) =\frac{ \ln 3 +\ln a}{\ln4 + \ln a} \\
\end{eqnarray*}
\begin{eqnarray*}
xyz +1 = \frac{ \ln a}{ \ln4 +\ln a} +1 =\frac{ \ln 4+ 2 \ln a}{\ln 4 + \ln a} =2 \frac{\ln 2 + \ln a}{ \ln 4 + \ln a} =2yz.
\end{eqnarray*}
A: Like Donald, I would express $a,b,c$ in terms of $\ln2,\ln3,\ln a$ 
and eliminate $\ln2,\ln3,\ln a$. 
I think this is how the problem came into being.
$$x= \log_{2a}(a) =\frac{ \ln a}{\ln2 + \ln a}\iff x\ln2+(x-1)\ln a=0 \  \ \ \  (1)$$
$$y= \log_{3a}(2a) =\frac{\ln 2 + \ln a}{\ln3 + \ln a}\iff\ln 2+(1-y)\ln a -y\ln3=0\  \ \ \  (2)$$
$$z= \log_{4a}(3a) =\frac{ \ln 3 +\ln a}{2\ln2 + \ln a}\iff2z\ln2+(z-1)\ln a-\ln3=0\  \ \ \  (3)$$
Method $\#1:$
Solve $(2),(3)$ for $\ln2,\ln a$ in terms of $\ln3$
Put these values in $(1)$
Method $\#2:$
Using Cramer's Rule,
$$\begin{vmatrix}
x & x-1 & 0 \\ 
1 & 1-y & -y \\
2z & z-1 & -1
\end{vmatrix}=0$$
