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Logarithm

Logarithm

MODULES

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Basics of Logarithm
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Exponential Function
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Log Properties
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Applying Log Properties
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Characteristic & mantissa
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Additional Problems
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Logarithms 1
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PRACTICE

Logarithm : Level 1
Logarithm : Level 2
Logarithm : Level 3
ALL MODULES

CAT 2025 Lesson : Logarithm - Exponential Function

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2. Exponential Function

For any real number
xxx, ex=∑n=0∞xnn!e^{x} = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^{n}}{n!}ex=n=0∑∞​n!xn​ =x00!+x11!+x22!+x33!+...= \dfrac{x^{0}}{0!} + \dfrac{x^{1}}{1!} + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + ...=0!x0​+1!x1​+2!x2​+3!x3​+...

ex=1+x+x22!+x33!+...e^{x} = 1 + x + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + ...ex=1+x+2!x2​+3!x3​+...

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.

111) e0=1e^{0} = 1e0=1
222) e=1+1+12!+13!+...∼2.71828e = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + ... \thicksim 2.71828e=1+1+2!1​+3!1​+...∼2.71828
333) eee is to be treated as a constant in arithmetic operations.
e2×e3=e5e^{2} \times e^{3} = e^{5}e2×e3=e5
(e5)10=e50(e^{5})^{10} = e^{50}(e5)10=e50
e2e3=e−1\dfrac{e^{2}}{e^{3}} = e^{-1}e3e2​=e−1
444) Given eee is a positive constant, where xxx is a real number (positive or negative) exe^{x}ex is always positive, i.e. ex>0e^{x} > 0ex>0

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



Example 3

What is the value of loge51!+(loge5)22!+(loge5)23!+...?\dfrac{\mathrm{log} _{e} {5}}{1!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{2!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{3!} + ... ?1!loge​5​+2!(loge​5)2​+3!(loge​5)2​+...?

Solution

loge51!+(loge5)22!+(loge5)23!+...\dfrac{\mathrm{log} _{e} {5}}{1!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{2!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{3!} + ...1!loge​5​+2!(loge​5)2​+3!(loge​5)2​+...

=
(1+loge51!+(loge5)22!+(loge5)23!+...)−1\left(1 + \dfrac{\mathrm{log} _{e} {5}}{1!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{2!} + \dfrac{(\mathrm{log} _{e} {5})^{2}}{3!} + ...\right) - 1(1+1!loge​5​+2!(loge​5)2​+3!(loge​5)2​+...)−1 =eloge5−1=e^{\mathrm{log} _{e} {5}} - 1=eloge​5−1

=5−1=4= 5 - 1 = 4=5−1=4

Answer:
444

Note: The following is the proof for
eloge5=5e^{\mathrm{log} _{e} {5}} = 5eloge​5=5

Let
x=eloge5x = e^{\mathrm{log} _{e} {5}}x=eloge​5

Expressing
xxx as logarithm, we get
logex=loge5\mathrm{log} _{e} {x} = \mathrm{log} _{e} {5}loge​x=loge​5 ⇒ x=5x = 5x=5

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