logarithmic problem university entrance test

Question

I have a question regarding logarithm topic, this has bothered me for days and I could not figure things out. Here's the problem:

If $ \log_{2n}\ 2016 \ = \log_n 504\sqrt{2} $, then $n^6$ must be equal to...

I have figured several way out but still dont find the whole picture of it, it is pointless and it does not satisfy.

Answer 1

Hint: write it as follows, then solve the linear equation in $\ln n$:

$$ \begin{align} \log_{2n} 2016 = \log_n 504\sqrt{2} & \quad \iff \quad \frac{\ln2016}{\ln 2n} = \frac{\ln504\sqrt{2}}{\ln n} \\ & \quad \iff \quad \frac{\ln2016}{\ln 2 + \ln n} = \frac{\ln 2016 - \frac{3}{2}\,\ln 2 }{\ln n} \end{align} $$

Answer 2

We have $$\log_{2n} 2016 =\log_{n} 504\sqrt {2} $$ $$\Rightarrow \frac{\log 2016}{\log 504\sqrt {2}} =\frac {\log 2n}{\log n} $$ $$\Rightarrow \frac {\log 2016}{\log 2016-1.5\log 2} =\frac {\log 2+\log n}{\log n} $$ $$\Rightarrow \log n (\log 2016) = \log 2 (\log 2016) -1.5 (\log 2)^2 +\log n (\log 2016) -1.5 (\log 2)\log n $$ Hope you can take it from here.

Answer 3

We set them both as x. So we have the following by definition of log : $$(2n)^x=2016$$ $$n^x=504\sqrt2$$ Dividing those two equations you have $$2^x=4/\sqrt2 => 2^x=2\sqrt2 => \sqrt2 ^{2x} = \sqrt2 ^3 => x=3/2 $$ so $$n^x=504\sqrt2 => n^{3/2}=504\sqrt2 => n^6=504^4 4$$