One way to think about a floating point number is that what a unit represents can change. Together with the implied "1." of the mantissa, there are 53 bits to specify the binary mantissa. In other words, we are using 53 bits to specify the quantity of some power of 2. This point of view is particularly helpful when dealing with exceedingly large values that cannot be represented as a 64-bit unsigned integer, or very small values. However, it is also true for values that are within the range of what can be represented by integers. For example, let us consider the quantity of 3. We are trying to figure out LaTeX: 3=x \cdot 2^p3 = x 2 p where LaTeX: 2^{52} \le x < 2^{53}-12 52 x < 2 53 1, and LaTeX: pp is whatever value is needed to make this equation work. In this case, LaTeX: x=2^{52}+2^{51}x = 2 52 + 2 51, which means LaTeX: p=-51p = 51 because LaTeX: (2^{52}+2^{51})\cdot 2^{-51} = 2^1+2^0 = 3( 2 52 + 2 51 ) 2 51 = 2 1 + 2 0 = 3. Given that LaTeX: v=v =4.686e28, find the value of LaTeX: pp such that LaTeX: (v=x \cdot 2^p) \wedge (2^{52} \le x < 2^{53}-1)( v = x 2 p ) ( 2 52 x < 2 53 1 ).