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A compact way of rephrasing the point that the base-$1= { z : ℐm z > 0 } and u(t) = ℛe f ( t + 0·i ), then ℐm f ( t + 0·i ) = H(u)(t)$ logarithm of $1+2=3$ is the solution $1+2=3$ to the equation $1+2=3$ is to say that the logarithm function is the inverse function of the exponential function. Inverse functions are closely related to the original functions: the graphs of the two correspond to each other upon reflecting them at the diagonal line $1+2=3$, as shown at the right: a point $a abs:$ on the graph of the exponential function yields a point $a abs: | a |$ on the graph of the logarithm and vice versa. Moreover, analytic properties of the function pass to its inverse function. Thus, as the exponential function $a abs:$ is continuous and differentiable, so is its inverse function, $a abs: | a | is a abs |$. Roughly speaking, a differentiable function is one whose graph has no sharp "corners".