# 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: $$xyz+1=2yz$$

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!

For $a>0$, $a\neq\frac{1}{2}$, $a\neq\frac{1}{3}$ and $a\neq\frac{1}{4}$ we obtain: $$2yz-xyz=\frac{2\ln2a\ln3a}{\ln3a\ln4a}-\frac{\ln{a}\ln2a\ln3a}{\ln2a\ln3a\ln4a}=$$ $$=\frac{2\ln2a}{\ln4a}-\frac{\ln{a}}{\ln4a}=\frac{\ln\frac{4a^2}{a}}{\ln4a}=1.$$

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

$$2yz = 2\frac{\log(2a)}{\log(3a)} \frac{\log(3a)}{\log(4a)} = 2 \frac{\log(2a)}{\log(4a)} = \frac{\log(4a^2)}{\log(4a)} = \frac{\log(a) + \log(4a)}{\log(4a)} = \frac{\log(a)}{\log(4a)} + 1 = xyz+1.$$

\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*}

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,

$$\begin{vmatrix} x & x-1 & 0 \\ 1 & 1-y & -y \\ 2z & z-1 & -1 \end{vmatrix}=0$$