Quantile function of log-normal distribution The quantile function for log-normal distribution is given by 
$$F^{-1}(p)=\exp(\mu+\sigma\Phi^{-1}(p)),$$
where $0<p<1$ and $\Phi(p)$ is the CDF of a normal distribution.
I am trying to derive $F^{-1}(p)$ and arrive at the solution above.
Let $X$ be log-normally distributed and $Z\sim N(\mu,\sigma^2)$. Now
$$F(x)=\Phi\left(\frac{\ln x-\mu}{\sigma}\right)=\frac{1}{\sqrt{2\pi}}\int^{\frac{\ln x-\mu}{\sigma}}_{0}\exp \left(-\frac{1}{2} \left(\frac{z-\mu}{\sigma}\right)^2\right)dz.$$
Now let's solve it in terms of $x$.
$$x=\frac{1}{\sqrt{2\pi}}\int^{\frac{\ln x-\mu}{\sigma}}_{0}\exp \left(-\frac{1}{2} \left(\frac{z-\mu}{\sigma}\right)^2\right)dz \iff \\
x\sqrt{2\pi}=\int^{\frac{\ln x-\mu}{\sigma}}_{0}\exp \left(-\frac{1}{2} \left(\frac{z-\mu}{\sigma}\right)^2\right)dz \iff \\
x\sqrt{2\pi}=\exp \left(-\frac{1}{2} \left(\frac{\frac{\ln x-\mu}{\sigma}-\mu}{\sigma}\right)^2\right)-\exp \left(-\frac{1}{2} \left(\frac{0-\mu}{\sigma}\right)^2\right) \iff \\
x\sqrt{2\pi}=\exp \left(-\frac{1}{2} \left(\ln x -\mu -\sigma\mu\right)^2\right)-\exp \left(\frac{1}{2} \left(\frac{\mu}{\sigma}\right)^2\right) \iff \\
x\sqrt{2\pi}+\exp \left(\frac{1}{2} \left(\frac{\mu}{\sigma}\right)^2\right)=\exp \left(-\frac{1}{2} \left(\ln x -\mu -\sigma\mu\right)^2\right) \iff \\
\log\left(x\sqrt{2\pi}+\exp \left(\frac{1}{2} \left(\frac{\mu}{\sigma}\right)^2\right)\right)=-\frac{1}{2} \left(\ln x -\mu -\sigma\mu\right)^2$$
I am not sure how to go on from here.
 A: Let $\log (X) \sim \mathcal{N}(\mu,\sigma^2)$, than indeed
$$
  F(x) = \Phi\left( \frac{\log(X) - \mu}{\sigma} \right).
$$
We know that $F(x) \in [0,1]$, so let $F(x) = p$, than we see that
$$
  p = \Phi\left( \frac{\log(F^{-1}(p)) - \mu}{\sigma} \right).
$$
Solving for the quantile function $F^{-1}(p)$ gives us
$$
  \Phi^{-1}(p) = \frac{\log(F^{-1}(p)) - \mu}{\sigma} \iff\\
  \sigma \Phi^{-1}(p) = \log(F^{-1}(p)) - mu \iff\\
  \mu + \sigma\Phi^{-1}(p) = \log(F^{-1}(p)) \iff\\
  F^{-1}(p) = \exp(\mu + \sigma\Phi^{-1}(p)).
$$
A: Start from the definition $$P(X \geq m_p)=1-p$$
where $m_p$ is your $F^{-1}(p).$
Using the fact that the $\log$ function is increasing
$$P(log(X) \geq log(m_p))=1-p$$
We have that $Z:=log(X)\sim \mathcal{N}(\mu,\sigma^2)$
$$P(\mu+\sigma Y \geq log(m_p))=1-p$$
where $Y \sim \mathcal{N}(0,1)$
Thus,
$$P(Y \geq \frac{log(m_p)-\mu}{\sigma})=1-p$$
$$1-P(Y \leq \frac{log(m_p)-\mu}{\sigma})=1-p$$
$$P(Y \leq \frac{log(m_p)-\mu}{\sigma})=p$$
$$\Phi\left(\frac{log(m_p)-\mu}{\sigma}\right)=p$$
where $\Phi$ is the CDF of a standard normal variable
Using the inverse of $\Phi$, we have 
$$\frac{log(m_p)-\mu}{\sigma}=\Phi^{-1}(p)$$
$$log(m_p)=\mu+\sigma \Phi^{-1}(p)$$
$$m_p=\exp\left(\mu+\sigma \Phi^{-1}(p)\right)$$
The result wanted
