Calculus

Limits and Continuity of Functions

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Natural Logarithms

Logarithms with base e, where e is an irrational number whose value is 2.718281828... 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 (a > 0, a ≠ 1), 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 \(1.\)

Natural and decimal logarithms
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.

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

See more problems on Page 2.

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