Natural Logarithms

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Logarithms with base \(e,\) where \(e\) is an irrational number whose value is \(2.718281828\ldots,\) are called natural logarithms. The natural logarithm of \(x\) is denoted by \(\ln x.\) Natural logarithms are widely used in mathematics, physics and engineering.

Relationship between natural logarithm of a number and logarithm of the number to base \(a\)

Let \(a\) be the base of logarithm \(\left({a \gt 0}\right.,\) \(\left. {a \ne 1}\right),\) and let

\[y = {\log _a}x.\]

This yields

\[{a^y} = x.\]

By taking the natural logarithm of both sides, we have

\[\ln {a^y} = \ln x,\;\; \Rightarrow y\ln a = \ln x,\;\; \Rightarrow y = \frac{1}{{\ln a}}\ln x,\;\; \Rightarrow {\log _a}x = \frac{{\ln x}}{{\ln a}}.\]

The last formula expresses logarithm of a number \(x\) to base \(a\) in terms of the natural logarithm of this number. By setting \(x = e,\) we have

\[{\log _a}e = \frac{1}{{\ln a}}\ln e = \frac{1}{{\ln a}}.\]

If \(a = 10,\) we obtain:

\[{\log _{10}}x = \lg x = M\,{\ln x} ,\;\;\; \text{where}\;\;M = \frac{1}{{\ln a}} = \lg e \approx 0.43429 \ldots \]

The inverse relationship is

\[\ln x = \frac{1}{M}\lg x,\;\;\; \text{where}\;\;\frac{1}{M} = \ln 10 \approx 2.30258 \ldots\]

Graphs of the functions \(y = \ln x\) and \(y = \lg x\) are shown in Figure \(1.\)

Solved Problems

Click or tap a problem to see the solution.

Example 1

Calculate \[\ln {\frac{1}{{\sqrt e }}}.\]

Example 2

Write as one logarithm:

\[{\frac{1}{3}}\ln \left( {x - 1} \right) - {{\frac{1}{2}}\ln \left( {x + 1} \right)} + {2\ln x}.\]

Example 1.

Calculate \[\ln {\frac{1}{{\sqrt e }}}.\]

Solution.

\[\ln \frac{1}{{\sqrt e }} = \ln {e^{ - \frac{1}{2}}} = - \frac{1}{2}\ln e = - \frac{1}{2}.\]

Example 2.

Write as one logarithm:

\[{\frac{1}{3}}\ln \left( {x - 1} \right) - {{\frac{1}{2}}\ln \left( {x + 1} \right)} + {2\ln x}.\]

Solution.

\[\frac{1}{3}\ln \left( {x - 1} \right) - \frac{1}{2}\ln \left( {x + 1} \right) + 2\ln x = \ln {\left( {x - 1} \right)^{\frac{1}{3}}} - \ln {\left( {x + 1} \right)^{\frac{1}{2}}} + \ln {x^2} = \ln \sqrt[3]{{x - 1}} - \ln \sqrt {x + 1} + \ln {x^2} = \ln \frac{{{x^2}\sqrt[3]{{x - 1}}}}{{\sqrt {x + 1} }}.\]