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MIT18_014F10_pset8sols

# MIT18_014F10_pset8sols - 18.014 Problem Set 8 Solutions...

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0 18.014 Problem Set 8 Solutions Total: 24 points Problem 1: Compute 1 xf �� (2 x ) dx 0 given that f �� is continuous for all x , and f (0) = 1, f (0) = 3, f (1) = 5, f (1) = 2, f (2) = 7, f (2) = 4. Solution (4 points) Applying integration by parts (theorem 5.5), we have 1 xf (2 x ) 1 1 1 1 1 1 1 1 xf �� (2 x ) dx = f (2 x ) dx = f (2) f (2) + f (0) = 0 . 2 2 2 4 4 2 0 We can use this theorem because x is differentiable with constant derivative 1 that is continuous and never changes sign, and f �� (2 x ) is continuous by hypothesis. Problem 2: Use the definition a x = e x log a to derive the following properties of general exponentials: (b) ( ab ) x = a x b x . (c) a x a y = a x + y . (d) ( a x ) y = ( a y ) x = a xy (e) Suppose a > 0, a = 1. Then y = a x if and only if x = log a y . Solution (4 points) (b) By the definition of the exponential function, part (ii) of theorem 3 of course notes M, part (i) of theorem 2 of course notes M, and the definition of the exponential function, we have ( ab ) x = e x log( ab ) = e x log( a )+ x log( b ) = e x log( a ) e x log( b ) = a x b x . (c) By the definition of the exponential function, part (i) of theorem 2 of course notes M, and the definition of the exponential function, we have x y x log( a ) y log( a ) ( x + y ) log( a ) x + y a a = e e = e = a . (d) After twice using the definition of the exponential function, using that the ex- ponential function and the logarithmic function are inverses, and again using the definition of the exponential function, we obtain ( a x ) y = e y log( a x ) = e y log( e x log a ) = e yx log( a ) = a xy . 1

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The same argument with the roles of x and y interchanged yields ( a y ) x = e x log( a y ) = e x log( e y log a ) = e xy log( a ) = a xy .
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