Question

How do you simplify $ {e^{ - 2\ln 5}} $ ?

Answer 1

Hint: An exponent refers to the number of times a number is multiplied by itself. For example, $ 2 $ to the $ 3rd $
 $ 2 \times 2 \times 2 = 8 $
 When an exponent is a negative number, the result is always a fraction. Fractions consist of a numerator over a denominator. In this distance, the numerator is always $ 1 $ . To find the denominator, pretend that the negative exponent is positive, and raise the number to that power, like this-
 $
  {a^{ - m}} = \dfrac{1}{{{a^m}}} \\
  {6^{ - 3}} = \dfrac{1}{{{6^3}}} \;
 $

Complete step by step solution:
We have,
 $ {e^{ - 2\ln 5}} $
Using the following properties of logarithm and exponentials
 $ n.\ln \left( m \right) = \ln \left( {{m^n}} \right) $ Here, Put $ n = - 2 $ and $ m = 5 $
 $ {e^{\ln \left( a \right)}} = a $ Here, Put $ a = {5^{ - 2}} $
We are going to get that.
During the initial simplification the log becomes $ {\log _e} $ of $ e $ , which has the value of
 $ {\log _e} = \ln e = 1 $
By definition the $ {\log _a}a = 1 $ whatever $ a $ is
(as long as $ a \ne 0 $ an $ a \ne 1 $ )
What $ {\log _a}x $ means is:
 What exponent do I use on $ a $ to get $ x $ ?
Example: $ {\log _{10}}1000 = 3 $ because $ {10^3} = 1000 $
So $ {\log _{10}}10 = 1 $ because $ {10^1} = 10 $
And this goes for any $ a $ in $ {\log _a} $ $ a $ because $ {a^1} = a $
In the same way, we will get
\[ = {e^{\ln \left( {{5^{ - 2}}} \right)}} = {5^{ - 2}} = \dfrac{1}{{{5^2}}} = \dfrac{1}{{25}} = 0.04\]
Hence, we have simplified the given expression.
So, the correct answer is “0.04”.

Note: A logarithm is the opposite of a power. In the other words, if we take a logarithm of a number, we undo an exponentiation. In other words, the logarithm gives the exponent as the output if you give it the exponentiation result as the input.
Basic rules for logarithms:
I.The product rule
II.The quotient rule
III.Log of a power
IV.Log of $ e $
V.Log of reciprocal
While applying the above rules, we have to be careful of what we are interchanging or what is going to base and what is becoming the super-script as it completely changes the resultant solution.