Nelson Jennings

2022-07-16

How to show $\mathrm{log}a\le n(\sqrt[n]{a}-1)\le \sqrt[n]{a}\mathrm{log}a$

Let $b=\sqrt[n]{a}$

How to show: $\mathrm{log}a\le n(\sqrt[n]{a}-1)\le \sqrt[n]{a}\mathrm{log}a$?

Thank you ;)

Let $b=\sqrt[n]{a}$

How to show: $\mathrm{log}a\le n(\sqrt[n]{a}-1)\le \sqrt[n]{a}\mathrm{log}a$?

Thank you ;)

tiltat9h

Beginner2022-07-17Added 14 answers

Noting $a={b}^{n}$, this is equivalent to

$n\mathrm{log}b\le n(b-1)\le nb\mathrm{log}b,$

i.e. just

To get this, employ the mean value theorem separately for $b\ge 1$ and $0<b<1$ (I assume the condition $b>0$ as $\mathrm{log}$ isn't defined for non-positive numbers). Notice first it does work for $b=1$, as all 3 quantities are $0$. If $b>1$, applying the mean value theorem to $x\mapsto \mathrm{log}x$ gives us that there exists $c\in [1,b]$ such that

$\mathrm{log}b-\mathrm{log}1=\frac{1}{c}(b-1),$

then noting $\mathrm{log}1=0$, and using the facts $c\in [1,b]$ and $b-1$ are positive gives us

$\frac{1}{b}(b-1)\le \frac{1}{c}(b-1)\le b-1$

gives us the result we want after a little bit of re-arranging. For $b<1$, the working is pretty much the same, except we get the last inequality from the fact $c\in [b,1]$ and $b-1$ is negative instead.

$n\mathrm{log}b\le n(b-1)\le nb\mathrm{log}b,$

i.e. just

To get this, employ the mean value theorem separately for $b\ge 1$ and $0<b<1$ (I assume the condition $b>0$ as $\mathrm{log}$ isn't defined for non-positive numbers). Notice first it does work for $b=1$, as all 3 quantities are $0$. If $b>1$, applying the mean value theorem to $x\mapsto \mathrm{log}x$ gives us that there exists $c\in [1,b]$ such that

$\mathrm{log}b-\mathrm{log}1=\frac{1}{c}(b-1),$

then noting $\mathrm{log}1=0$, and using the facts $c\in [1,b]$ and $b-1$ are positive gives us

$\frac{1}{b}(b-1)\le \frac{1}{c}(b-1)\le b-1$

gives us the result we want after a little bit of re-arranging. For $b<1$, the working is pretty much the same, except we get the last inequality from the fact $c\in [b,1]$ and $b-1$ is negative instead.

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