# Calculus

## Limits and Continuity of Functions # Natural Logarithms

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:

${{\log _{10}}x = \lg x = M\,{\ln x} ,\;\;\;}\kern-0.3pt {\text{where}\;\;M = \frac{1}{{\ln a}} = \lg e} \approx {0.43429 \ldots }$

The inverse relationship is

${\ln x = \frac{1}{M}\lg x,\;\;\;}\kern-0.3pt {\text{where}\;\;\frac{1}{M} }={ \ln 10 }\approx {2.30258 \ldots }$

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 $$\ln {\large\frac{1}{{\sqrt e }}\normalsize}.$$

### Example 2

Write as one logarithm: $${\large\frac{1}{3}\normalsize}\ln \left( {x – 1} \right) – {{\large\frac{1}{2}\normalsize}\ln \left( {x + 1} \right)} + {2\ln x}.$$

### Example 3

Sketch the graph of the function $$y = \ln \left( {x + 1} \right) – 1.$$

### Example 4

Sketch the graph of the function $$y = \left| {\ln x} \right|.$$

### Example 5

Sketch the graph of the function $$y = \left| {\ln \left| x \right|} \right|.$$

### Example 1.

Calculate $$\ln {\large\frac{1}{{\sqrt e }}\normalsize}.$$

Solution.

${\ln \frac{1}{{\sqrt e }} = \ln {e^{ – \large\frac{1}{2}\normalsize}} }={ – \frac{1}{2}\ln e = – \frac{1}{2}.}$

### Example 2.

Write as one logarithm: $${\large\frac{1}{3}\normalsize}\ln \left( {x – 1} \right) – {{\large\frac{1}{2}\normalsize}\ln \left( {x + 1} \right)} + {2\ln x}.$$

Solution.

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

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Problems 1-2
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Problems 3-5