If X is log-normal, is: $\frac{a}{\sqrt{b+cX}}$? I am working for the first time with log-normal distributions and I want to verify whether the following statement is true. I am not sure whether all the properties of the log-normal distribution hold also for a 'shifted log-normal distribution'.
If $X$ is a log-normal distribution:
$$X\sim \ln\mathcal{N}(\mu,\sigma^2)$$
Then does it follow that for $a,b,c>0$:
$$\exists\mu',\sigma':\frac{a}{\sqrt{b+cX}}\sim \ln\mathcal{N}(\mu',\sigma'^2)$$  
with:
$$\mu'=-\frac{\mu+\ln c+b}{2}$$  $$\sigma'^2=\frac{\sigma^2}{4}$$

For $t\ge 0$
$$P\left\{\frac{a}{\sqrt{b+cX}}\le t\right\}=P\left\{\frac{a^2}{b+cX}\le t^2\right\}=P\left\{X\ge\frac{(a/t)^2-b}{c}\right\}=\int_{\frac{(a/t)^2-b}{c}\vee0}^\infty \frac{1}{x\sigma\sqrt{2\pi}}\exp\left\{-\frac{(\ln(x)-\mu)^2}{\sigma^2}\right\}dx$$
Then for $t\in\left[0,\frac{a}{\sqrt{b}}\right]$
$$f_{X'}(t)=\frac{2a^2}{t^3}\frac{1}{((a/t)^2-b)\sigma\sqrt{2\pi}}\exp\left\{-\left(\ln\left(\frac{(a/t)^2-b}{c}\right)-\mu\right)^2\sigma^{-2}\right\}$$
And if $b=0$,
$$f_{X'}(t)=\frac{1}{t(\sigma/2)\sqrt{2\pi}}\exp\left\{-\frac{(\ln(t)-0.5[ln(a^2/c)-\mu])^2}{(\sigma/2)^2}\right\}$$
which is lognormal with parameters $\mu'=[ln(a^2/c)-\mu]/2$ and $\sigma'=\sigma/2$.
