the difference between these two logarithm

Question

I was just wondering what is the difference between ${1\over \ln(n^2)}$ and ${1\over \ln^2(n)}$

I know that ${1\over \ln(n^2)}$ is ${1\over 2\ln(n)}$ through the power rule, but I am not so sure about ${1\over \ln^2(n)}$.

Answer 1

Note that:

$$\ln^2(n)=\ln(n)\cdot\ln(n)$$

Therefore:

$$\frac{1}{\ln^2(n)}=\frac{1}{\ln(n)\cdot\ln(n)}$$

Whereas:

$$\frac{1}{2\ln(n)}=\frac{1}{\ln(n)+\ln(n)}$$

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They are equal if and only if $\ln(n)=2$, or in other words, if and only if $n=e^2$.

Answer 2

Not a good example: $$ n = \mathrm e^2 \implies \frac{1}{\log n^2} = \frac14\qquad \frac1{\log^2n} = \frac14 $$ Good example: $$ n = \mathrm e^3 \implies \frac{1}{\log n^2} = \frac16\qquad \frac1{\log^2n} = \frac19 $$

Answer 3

Let's say $x = \frac{1}{\ln n}$, then $$\frac{1}{\ln n^2} = \frac{1}{2\ln n} = \frac{x}{2}$$ $$\frac{1}{\ln^2 n} = (\frac{1}{\ln n})^2 = x^2$$

You can clearly notice the difference between $\frac{x}{2}$ and $x^2$. Hope it cleared your doubt.