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## julie 3 years ago How exactly do these procedures work? (define (square x) (exp (double (log x)))) (define (double x) (+ x x)) Delete Cancel Submit

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1. Owlfred
• 3 years ago
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• 3 years ago
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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.

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