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Proof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−y = ex · 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ N.

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T~ 2 ˘Geom(1=m) be the time to get the remaining chunk The second random variable is geometric by the memoryless property of the geometric distribution Our total time to collect the rst two movies is T~ = T~ 1 T~ 2 so by linearity of expectation and the mean of a geometric random variable, ET~ = ET~ 1 ET~ 2 = m=2 m = 3=2 m (c)(. In C, there is shortcircuiting, so the statement y=z will not be evaluated until x becomes zero When x == 0, since z also decrements the same way, z == 0Hence y will also be zero at that time due to the assignment The statement y=z also returns y at this point which will be evaluated as a condition, and since that is also 0, the else break will be hit.