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julie
How exactly do these procedures work? (define (square x) (exp (double (log x)))) (define (double x) (+ x x))
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Pretty staightforward; remember how logs/exponents work: \[\begin{align} exp(double (log x)) &= e^{2 \ln x}\\ &= e^{\ln x^2}\\ &= x^2 \end{align}\] Have a look at http://en.wikipedia.org/wiki/Logarithm for more.