# Derive mode of lognornormal distribution

I need to generate some random data from lognormal distribution, where I set the mode and standart deviation of that lognormal distribution. For this purpose I choose to use random numbers generator from lognormal distribution. This generator takes two numbers, that are mean and sd of underlying normal distribution.

So far its clear I need to derive mean and sd of normal distribution, which is underlaying for lognormal distribution where I know mode and sd. I know the equations for derivation of mean and sd:

NOTATION:

n(x) = mean of normal distribution

sd(x) = sd of normal distribution

n(y) = mean of lognormal distribution

sd(y) = sd of lognormal distribution

mode(y) = mode of lognormal distribution

EQUATIONS:

$$n(x) = 2*ln(n(y)) - (1/2)*ln(sd(y)^2 + n(y)^2)$$ $$sd(x) = -2*ln(n(y)) + ln(sd(y)^2 + n(y)^2)$$ $$mode(y) = exp(n(y) - sd(y)^2)$$

Here I stuck because I cant get the equation for $n(y)$ from these equations, that I need to compute $n(x)$. So far I ended:

$$mode(y) = exp(4*ln(n(y))-3/2*ln(n(y)^2 - sd(y)^2))$$ $$mode(y)^{2/3}*sd(y)^2 = n(y)^2 * (n(y)^{2/3} - mode(y)^{2/3})$$

Can anybody help me to complete this derivation? Or there are any other ways to anallytically derive mean of lognormal distribution from mode and sd of lognormal distribution?

If the underlying normal distribution has mean $\mu$ and variance $\sigma^2$ then the lognormal has mean $\exp(\mu+\sigma^2/2)$, mode $\exp(\mu - \sigma^2)$, and variance $(\exp(\sigma^2) -1)\exp(2\mu+\sigma^2)$. You say you set the mode and standard deviation of the lognormal, let's say as $m$ and $s$
Then $m=\exp(\mu - \sigma^2)$ so $\log_e(m) = \mu - \sigma^2$ and $$\mu = \log_e(m) + \sigma^2$$ leaving us to find $\sigma$. Meanwhile $s^2 = (\exp(\sigma^2) -1)\exp(2\mu+\sigma^2)$ which with some substitution and manipulation gives $$(\exp(\sigma^2))^4 - (\exp(\sigma^2))^3- \dfrac{s^2}{m^2} =0$$ which is a quartic equation in $\exp(\sigma^2)$ with what looks like one root greater than $1$ when $\frac{s^2}{m^2} \gt 0$ and so it may be possible to solve for $\sigma$ even if unattractive in a closed form