I have a random variable $X$ where $$10\log_{10}(X)\sim N(\mu, \sigma^2)$$. So we know that $X^{\frac{10}{\ln{10}}}\sim LN(\mu,\sigma^2)$ and therefore I can conclude that $E[X^{\frac{10}{\ln{10}}}]=\exp(\mu+\frac{\sigma^2}{2})$. Is it right? Further I want to know how we can find $E[X^{-1}]$? Any help in this regard will be much appreciated. Thanks in advance.
I have been able to find the expectation of $X^{-1}$. The steps are given below.
1- We know that $X^{\frac{10}{\ln{10}}}\sim LN(\mu,\sigma^2)$. So introduce a new variable $X^{\frac{10}{\ln{10}}}=\exp(\mu+\sigma Z)$ where $Z$ is astandard normal random variable.
2- Now take $-\frac{\ln(10)}{10}$ power of both sides.
3- After that take the expectation of both sides and complete the squares and use the fact that integral of pdf is 1 overall its possible range.
4- The final answer will be $$10^{\frac{-\mu+(\ln(10)/10)(\sigma^2/2)}{10}}$$. Thank you.
Your observation $X^{\frac{10}{\ln10}}\sim\mathsf{Lognormal}(\mu,\sigma^2)$ is correct.
If $Z\sim\mathsf{Lognormal}\left(\mu,\sigma^{2}\right)$ then we can write $Z=e^{\mu+\sigma U}$ where $U$ has standard normal distribution.
Now observe that for $a>0$ we have $Z^{a}=e^{a\mu+a\sigma U}\sim\mathsf{Lognormal}\left(a\mu,a^{2}\sigma^{2}\right)$.
If $a<0$ then $Z^{a}=e^{a\mu+a\sigma U}=e^{a\mu+\left(-a\right)\sigma\left(-U\right)}\sim\mathsf{Lognormal}\left(a\mu,a^{2}\sigma^{2}\right)$ since also $-U$ has standard normal distribution.
So in general if $Z\sim\mathsf{Lognormal}\left(\mu,\sigma^{2}\right)$ and $a\neq0$ then $Z^{a}\sim\mathsf{Lognormal}\left(a\mu,a^{2}\sigma^{2}\right)$
Applying this you can find that $X^{-1}\sim\mathsf{Lognormal}\left(-\frac{\mu\ln10}{10},\left(\frac{\sigma\ln10}{10}\right)^{2}\right)$.
Then:$$\mathbb{E}X^{-1}=e^{-\frac{\mu\ln10}{10}+\frac{1}{2}\left(\frac{\sigma\ln10}{10}\right)^{2}}$$