Solveeit Logo
Login

Question

Question: How do you simplify \( {e^{ - 2\ln 5}} \) ?...

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

Explanation

Solution

Hint : An exponent refers to the number of times a number is multiplied by itself. For example, 22 to the 3rd3rd
2×2×2=82 \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 11 . To find the denominator, pretend that the negative exponent is positive, and raise the number to that power, like this-
a−m=1am 6−3=163   {a^{ - m}} = \dfrac{1}{{{a^m}}} \\\ {6^{ - 3}} = \dfrac{1}{{{6^3}}} \;

Complete step by step solution:
We have,
e−2ln⁡5{e^{ - 2\ln 5}}
Using the following properties of logarithm and exponentials
n.ln⁡(m)=ln⁡(mn)n.\ln \left( m \right) = \ln \left( {{m^n}} \right) Here, Put n=−2n = - 2 and m=5m = 5
eln⁡(a)=a{e^{\ln \left( a \right)}} = a Here, Put a=5−2a = {5^{ - 2}}
We are going to get that.
During the initial simplification the log becomes log⁡e{\log _e} of ee , which has the value of
log⁡e=ln⁡e=1{\log _e} = \ln e = 1
By definition the log⁡aa=1{\log _a}a = 1 whatever aa is
(as long as a≠0a \ne 0 an a≠1a \ne 1 )
What log⁡ax{\log _a}x means is:
What exponent do I use on aa to get xx ?
Example: log⁡101000=3{\log _{10}}1000 = 3 because 103=1000{10^3} = 1000
So log⁡1010=1{\log _{10}}10 = 1 because 101=10{10^1} = 10
And this goes for any aa in log⁡a{\log _a} aa because a1=a{a^1} = a
In the same way, we will get
=eln⁡(5−2)=5−2=152=125=0.04= {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 ee
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.