I have seen this equation from a link named Asisten and German Academy, (it is a video of Facebook) where there is a complicate solution (I invite to watch it) for
$$(2n)^{\log 2}=(5n)^{\log 5}$$
I have adopted, instead, this approach:
$$(2)^{\log 2}(n)^{\log 2}=(5)^{\log 5}(n)^{\log 5} \iff 2n^{\log 2}=5n^{\log 5}$$
After
$$\frac{n^{\log 2}}{n^{\log 5}}=\frac 52 \iff n^{(\log 2-\log 5)}=\frac 52$$
$$\log(n^{(\log 2-\log 5)})= \log 5-\log 2 $$
$$(\log 2-\log 5)\log n=\log 5-\log 2 \iff \log n =-1$$
Hence $$e^{\log n}=e^{-1}\implies n=\frac 1e$$
Now I have seen the solution is $n=1/10$.
Is it different my solution why my base is $e$ and not $10$? Generally I have seen that $\log=\log_{10}$. In Italy we used often $\log=\log_e$. I yet thought with the old notation that $\operatorname{Log}=\log_{10}.$ I not adopted often $\ln$ where the base is neperian.
Best Answer
Regardless of what base you are using, it is not true that both $$ 2^{\log(2)} = 2 \qquad\text{and}\qquad 5^{\log(5)} = 5. $$ It is certainly true that $2^{\log_2(2)} = 2$, but the notation $\log$, without a subscript, does not mean "choose whatever base you like and go with it," it typically denotes a specific base.
In mathematics, $\log$ typically means the natural logarithm,[1] which has base $\mathrm{e}$. Under that assumption, $$ 2^{\log(2)} \approx 1.617 \qquad\text{and}\qquad 5^{\log(5)} \approx 13.334. $$ If $\log$ denotes the common logarithm, which as base $10$, then $$ 2^{\log(2)} \approx 1.232 \qquad\text{and}\qquad 5^{\log(5)} \approx 3.080. $$
In any event, assuming that $\log$ is the natural logarithm, $$\begin{align} &(2n)^{\log(2)} = (5n)^{\log(5)} \\ &\qquad\iff \frac{2^{\log(2)}}{5^{\log(5)}} = \frac{n^{\log(5)}}{n^{\log(2)}} \\ &\qquad\iff\frac{\mathrm{e}^{\log(2)^2}}{\mathrm{e}^{\log(5)^2}} = \frac{n^{\log(5)}}{n^{\log(2)}} \label{eq1}\tag{*} \\ &\qquad\iff \mathrm{e}^{\log(2)^2 - \log(5)^2} = n^{\log(5)-\log(2)} \\ &\qquad\iff \log(2)^2 - \log(5)^2 = \log(n) \left( \log(5)-\log(2) \right) \\ &\qquad\iff \log(n) = \frac{\log(2)^2 - \log(5)^2}{\log(5)-\log(2)} = -\left( \log(2)+\log(5)\right) = -\log(10) \\ &\qquad\iff n = \mathrm{e}^{-\log(10)} = \frac{1}{10}. \end{align}$$
At (\ref{eq1}), I am using the fact that for any positive real number $a$, we can write $$ a = \exp(\log(a)) = \mathrm{e}^{\log(a)}, $$ since the exponential and logarithm are inverse functions to each other. This implies that, for exmaple, $$ 2^{\log(2)} = \left( \mathrm{e}^{\log(2)} \right)^{\log(2)} = \mathrm{e}^{\log(2)\cdot \log(2)} = \mathrm{e}^{\log(2)^2}. $$
Note that the choice of the natural logarithm is irrelevant here. Replacing the natural logarithm with the logarithm with base $b$ will have the effect of replacing every occurrence of $\mathrm{e}$ with $b$ in the above computation, but this will not change the result.
[1] The question asks about "neperian" logarithms. This term does not exist in English, though it probably refers to the Napierian logarithm (also spelled "Naperian"), named for John Napier, the Scottish mathematician responsible for working out large tables of logarithm ("Napier's bones"). The Napierian logarithm is the natural logarithm.