What will be the value of n?

Question

$$ \log_{10}2 = \frac{1}{n} \times \log_{10}n\, $$

Can anybody please help me finding the value of n ?

Sorry for earlier typo..Edited the questions

Answer 1

This simplifies to $n=\log_{2}(n)$, so $2^{n}=n$. This equation cannot be solved algebraically but has a solution in terms of the Lambert W Function$W$:

$$n=-\frac{W_m(-\log(2))}{\log(2)}$$

Answer 2

HINT: $$\log_{10}2=\frac{1}{n}\times \log_{10}n$$ $$n\log_{10}2=\log_{10}n$$ $$\log_{10}2^{n}=\log_{10}n$$ $$2^{n}=n$$

Answer 3

Re-arranging your expression (after considering the typo), we have $$\log_{10}2=\frac{1}{n}\cdot \log_{10}n$$ or, $$n\log_{10}2=\log_{10}n$$ or, $$\log_{10}2^{n}=\log_{10}n$$ or, $$2^{n}=n$$

But the graphs of the equations $y=2^{x}$ and $y=x$ never intersect. So there are no real solutions for $n$ in this equation.

Answer 4

For real $n$, $$ \frac{\log(n)}{n}\le\frac1e\lt\log(2)\tag{1} $$ so there is no real solution.

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We can find complex solutions. If $$ \log(2)=\frac1n\log(n)\tag{2} $$ then $$ -\log(2)=\frac1n\log\left(\frac1n\right)=\log\left(\frac1n\right)e^{\log\left(\frac1n\right)}\tag{3} $$ Thus, if $$ n=e^{-W(-\log(2))}\tag{4} $$ for any branch of the Lambert W, then $$ \begin{align} \frac1n\log(n) &=\overbrace{-W(-\log(2))\vphantom{e^W}}^{\log(n)}\overbrace{e^{W(-\log(2))}}^{\frac1n}\\ &=-(-\log(2))\\[4pt] &=\log(2)\tag{5} \end{align} $$ Since $n=\frac{\log(n)}{\log(2)}$, we also have $$ n=\frac{-W(-\log(2))}{\log(2)}\tag{6} $$ Therefore, we can use either $(4)$ or $(6)$ to compute $n$.