Solved – transforming uniform to exponential distribution

You can change the density of a univariate continuous random variable $X$ into each by the following transform $$mathcal F_2^{-1}left(mathcal F_1(X)right),$$ where $mathcal F_1$ and $mathcal F_2$ are the cumulative distribution functions of your source density and your target density, respectively.

In your example, $mathcal F_1$ is the c.d.f. of the uniform distribution on an interval $[a,b]$, which is $$mathcal F_1(x) = frac{x-a}{b-a}.$$

The c.d.f. of an exponential distribution is $mathcal F_2(x) = 1 − e^{-lambda x}$ and its inverse is $$mathcal F_2^{-1}(y) = -frac{log(1-y)}{lambda}.$$ Putting it together, you get $$mathcal F_2^{-1}(y) = -frac{logleft(1-frac{x-a}{b-a}right)}{lambda}.$$

This has nothing to do with the Box-Muller transform that transforms uniform random variables on $[0,1]$ into a two-dimensional isotropic Gaussian.