# Beyond power and polynomial relations

## The logarithm function \(\normalsize y=\ln x\) and its inverse

We’ve already seen that the (natural) logarithm function \(\normalsize y=\ln x\) could be defined, at least approximately, as an area under the hyperbolic function \(\normalsize y=1/x\). And we’ve seen that purely on geometric grounds, using only the basic scaling associated with the hyperbola, that \[\Large \ln (ab)=\ln a + \ln b. \tag{1}\label{1}\]The inverse function to the logarithm is \(\normalsize y=e^x\), called the*exponential function*, which is defined by the condition that \[\Large y=e^x \quad \text{precisely when}\quad x=\ln y . \tag{2}\label{2}\]Geometrically the meaning is that the graph of the exponential function and the logarithm function are reflections of each other in the line \(\normalsize y=x\).An important property of the exponential function is that \[\Large e^0=1. \tag{3}\label{3}\]

Q1(E): Explain why this formula holds.

Q2(M): Using equations \((\ref{1})\) and \((\ref{2})\), prove that \((\ref{4})\) holds.

*Euler’s number*:

## Answers

A1.Recall from the definition that \(\normalsize \ln x\) is the area under the hyperbola \(\normalsize y=1/t\) from \(\normalsize t=1\) to \(\normalsize t=x\). So \(\normalsize \ln 1 = 0\) which is to say \(\normalsize e^0 = 1\).\[\Large \ln(ab) = \ln a + \ln b.\]A2.We use equations \(\normalsize (\ref{1})\) and \(\normalsize (\ref{2})\) in combination. Starting withExponentiate both sides\[\Large ab = e^{\ln a + \ln b}.\]Now let \(\normalsize a = e^x\) and \(\normalsize b = e^y\), so that\[\Large e^x e^y = e^{\ln(e^x) + \ln(e^y)} = e^{x+y}.\]

#### Maths for Humans: Linear, Quadratic & Inverse Relations

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