+91 9600 121 800

Back

Quant

Algebra

Logarithm

ALL MODULES

CAT 2025 Lesson : Logarithm - Exponential Function

2. Exponential Function

For any real number

x

e^{x} = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^{n}}{n!}

= \dfrac{x^{0}}{0!} + \dfrac{x^{1}}{1!} + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + ...

e^{x} = 1 + x + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + ...

While complex questions pertaining to the exponential function are not expected, there could be questions pertaining to basic properties. These listed below are derived from the above equation.

1

)

e^{0} = 1

2

)

e = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + ... \thicksim 2.71828

3

)

e

is to be treated as a constant in arithmetic operations.

e^{2} \times e^{3} = e^{5}

(e^{5})^{10} = e^{50}

\dfrac{e^{2}}{e^{3}} = e^{-1}

4

) Given

e

is a positive constant, where

x

is a real number (positive or negative)

e^{x}

is always positive, i.e.

e^{x} > 0

The graph of the exponential function and logarithmic functions are as below.

Example 3

What is the value of

\dfrac{\mathrm{log} _{e} {5}}{1!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{2!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{3!} + ... ?

Solution

\dfrac{\mathrm{log} _{e} {5}}{1!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{2!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{3!} + ...

\left(1 + \dfrac{\mathrm{log} _{e} {5}}{1!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{2!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{3!} + ...\right) - 1

=e^{\mathrm{log} _{e} {5}} - 1

= 5 - 1 = 4

Answer:

4

Note: The following is the proof for

e^{\mathrm{log} _{e} {5}} = 5

Let

x = e^{\mathrm{log} _{e} {5}}

Expressing

x

as logarithm, we get

\mathrm{log} _{e} {x} = \mathrm{log} _{e} {5}

⇒

x = 5

Loading Video....