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Last modified on March 23rd, 2024

The natural logarithm (base-e-logarithm) of a positive real number x, represented by lnx or log_{e}x, is the exponent to which the base ‘e’ (≈ 2.718…, Euler’s number) is raised to obtain ‘x.’

Mathematically, ln(x) = log_{e}(x) = y if and only if e^{y} = x

It is also written as:

${\ln x=\int ^{x}_{1}\dfrac{1}{t}dt}$

The number 2.718… is used naturally in math and science calculations, like pie (π) in geometry.

The geometric interpretation of the natural logarithmic function y = lnx is shown below.

Here, the natural logarithmic function has an asymptote at x = 0 and an x-intercept at (1, 0).

Now, let us verify the natural logarithm formula ${\ln x=\int ^{x}_{1}\dfrac{1}{t}dt}$

According to the fundamental theorem of calculus, if F(x) is an antiderivative of f(x) for the interval [a, b], then ${\int ^{b}_{a}f\left( x\right) dx=f\left( b\right) -f\left( a\right)}$

Now, let us find an antiderivative of ${f\left( t\right) =\dfrac{1}{t}}$

${\int \dfrac{1}{t}dt=\ln \left| t\right| +C}$, here c is the integrating constant.

Applying the fundamental theorem of calculus, we get

${\int ^{x}_{1}\dfrac{1}{t}dt=\ln \left| x\right| -\ln \left| 1\right|}$

= ${\ln \left| x\right| -0}$

= ${\ln \left| x\right|}$

Thus, ${\int ^{x}_{1}\dfrac{1}{t}dt=\ln \left| x\right|}$, is proved.

The natural logarithm follows all the properties of the logarithm.

ln(xy) = lnx + lny

**Example:** ln(10) = ln(5) + ln(2)

${\ln \left( \dfrac{x}{y}\right) =\ln x-\ln y}$

**Example:** ${\ln \left( \dfrac{42}{5}\right) =\ln 42-\ln 5}$

ln(x^{n}) = n lnx

**Example:** ln((5z)^{7}) = 7 ln(5z)

${\ln x=\log _{e}x=\dfrac{\log _{b}x}{\log _{b}e}}$

Or, lnx ⋅ log_{b}e = log_{b}x

**Example:** ${\ln \left( 2p^{2}\right) =\log _{e}\left( 2p^{2}\right) =\dfrac{\log _{b}\left( 2p^{2}\right) }{\log _{b}e}}$, for any base b

ln(e) = 1

ln(1) = 0

ln(∞) = ∞, which means while the argument ‘x’ approaches to infinity, the limit of ln(x) is:

lim ln(x) = ∞, when x → ∞

${\dfrac{d}{dx}\left( \ln \left( x\right) \right) =\dfrac{1}{x}}$

${\int \ln \left( x\right) dx=x\left( \ln \left( x\right) -1\right) +c}$

**ln of Negative Numbers:**ln of any negative number is undefined.**ln of ‘0’ and the Limit:**ln(0) is undefined, which means while the argument ‘x’ approaches zero, the limit of ln(0) is minus infinity. That is, lim ln(0) = -∞, when x → 0^{+}**ln of ‘e’ raised to the power ‘x’:**ln(e^{x}) = x**‘e’ raised to the ln power:**e^{ln(x)}= x

**Simplify: ln(e ^{4x}) + 4ln(x) – 2ln(2x)**

Solution:

Here, ln(e^{4x}) + 4ln(x) – 2ln(2x)

= 4x ln(e) + ln(x^{4}) – ln((2x)^{2}) (by the power rule of logarithm)

= 4x + ln(x^{4}) – ln(4x^{2}) (since ln(e) = 1)

= ${4x+\ln \left( \dfrac{x^{4}}{4x^{2}}\right)}$ (by the quotient rule of logarithm)

= ${4x+\ln \left( \dfrac{x^{2}}{4}\right)}$

= ${4x+\ln \left( \left( \dfrac{x}{2}\right) ^{2}\right)}$

= ${4x+2\ln \left( \dfrac{x}{2}\right)}$ (by the power rule of logarithm)

Thus, ln(e^{4x}) + 4ln(x) – 2ln(2x) = ${4x+2\ln \left( \dfrac{x}{2}\right)}$

**Convert the natural logarithmic equation ln(15) = 2.708 (corrected to 3 decimal places) into its exponential form.**

Solution:

As we know, ln(x) = y ⇔ e^{y} = x

Here, ln(15) = 2.708Thus, e^{2.708} = 15