1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

\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*}

$\endgroup$
0
$\begingroup$

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,

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.