I am looking for the answer of the $(\lg n)$-th root of $n$, that is, $\sqrt[\lg n]{n}$. What is the answer and what log property should I use here? Please assume base as $2$ and $n$ as a natural number.
What is the $(\lg n)$-th root of $n$?
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$\begingroup$
arithmetic
logarithms
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3Take a look at this nice reference: http://meta.math.stackexchange.com/q/5020/856 – 2012-08-30
4 Answers
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Take the $\log$ of this expression, to get $\log (n^{1/\log(n)}) = \frac{1}{\log(n)} \log(n) = 1.$ This means that: $n^{1/\log(n)}=2$ or generally, the base of your $\log$.
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Hint: Let $b$ be any positive number different from $1$. Then $x^y=b^{(y\,\log_b x)}$.
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0@Geek: there are lots of power rules. [Here is a link.](http://en.wikipedia.org/wiki/Logarithm#Product.2C_quotient.2C_power.2C_and_root) this follows easily from the definition of logarithm, and the general power law for exponentiation, $(p^q)^r=p^{(qr)}$. – 2012-08-30
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It's not a matter of applying power rules in sneaky ways. It directly follows from the definitions:
- By the definition of logarithm, $\log_2 n$ is the number $L$ such that $2^L=n$.
- By the definition of higher roots, $\sqrt[L]n$ is the (positive) number $R$ such that $R^L=n$.
So $L$ is explicitly defined such as to make sure the $L$th root of $n$ is $2$!
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$\sqrt[\log_2 n]{n}=\color{red}{n}^{1/\log_2 n}=\color{red}2^{\frac{\color{red}{\log_2 n}}{\log_2 n}}=2^1$