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Let $X$ be a Gaussian distributed random variable and $f: \mathbb{R} \to \mathbb{R}^+$ a transformation function. Looking for $f$ that makes $f(x)$ exponentially (or Laplace) distributed.
For instance if $f=x^2$, $f(x)$ is Chi-square distributed, if $f=e^x$, $f(x)$ becomes lognormally distributed, etc.
If $X$ is a random variable with continuous, strictly increasing CDF $F$, then $F(X)$ is uniformly distributed on $[0,1]$ (this is a really good exercise if you don't already know it). Conversely, if $U$ is uniformly distributed on $[0,1]$, then $F^{-1}(U)$ has the same distribution as $X$. This latter fact can be generalised, but then you need to define $F^{-1}$ properly.
Combining these two observations, if $Z\sim \mathcal{N}(0,1)$, then $\Phi(Z)$ is uniformly distributed on $[0,1]$, where $\Phi$ denotes the standard Gaussian CDF. Note that the CDF of the standard Laplace distribution is $$ F(x)= \begin{cases} \frac{1}{2} e^{x} & x<0 \\ \frac{1}{2}(2-e^{-x}) & x\geq 0 \end{cases} $$ Thus, $$ F^{-1}(u)=\begin{cases} \log(2u) & 0<u< \frac{1}{2} \\ -\log(2(1-u)) & \frac{1}{2}\leq u< 1 \end{cases} $$ and $F^{-1}(\Phi(Z))$ is Laplace distributed. You can find a transformation which yields an exponential distribution in a similar way.
