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Question: How do you simplify \(\dfrac{{\log 4}}{{\log 2}}\)?...

How do you simplify log⁡4log⁡2\dfrac{{\log 4}}{{\log 2}}?

Explanation

Solution

In order to determine the value of the above question ,use basic exponent property.
For Example , we can write 4=224 = {2^2}so that we can apply the logarithm property on this
expression.
The logarithmic property
nlog⁡m=log⁡mn mlog⁡(n)=n  n\log m = \log {m^n} \\\ {m^{\log (n)}} = n \\\
will be applied to the expression so that we can cancel out the common factor. After applying all these steps we will get our desired simplification.

Complete step by step solution:
Applying the above exponent rule in this question, we get:
log⁡4log⁡2\dfrac{{\log 4}}{{\log 2}}
=log⁡22log⁡2\dfrac{{\log {2^2}}}{{\log 2}}----eq1
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our expression.
But we also need to know that the logs of the same bases are actually allowed to cancel out each other.
First, we are going to rewrite and solve with the help of the following properties of natural
logarithms. By applying the logarithmic property
nlog⁡m=log⁡mn mlog⁡(n)=n  n\log m = \log {m^n} \\\ {m^{\log (n)}} = n \\\
=log⁡22log⁡2\dfrac{{\log {2^2}}}{{\log 2}}
=2∙log⁡2log⁡2\dfrac{{2 \bullet \log 2}}{{\log 2}}
= 2
Therefore ,the simplification of log⁡4log⁡2\dfrac{{\log 4}}{{\log 2}} is 2.

Note: 1. Value of the constant ”e” is equal to 2.71828.
2. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number , we actually undo an exponentiation.
3.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values .
log⁡b(mn)=log⁡b(m)+log⁡b(n){\log _b}(mn) = {\log _b}(m) + {\log _b}(n)
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
log⁡b(mn)=log⁡b(m)−log⁡b(n){\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
nlog⁡m=log⁡mnn\log m = \log {m^n}
6. The above guidelines work just if the bases are the equivalent. For example, the expression log⁡d(m)+log⁡b(n){\log_d}(m) + {\log _b}(n)can't be improved, on the grounds that the bases (the "d" and the "b") are not
the equivalent, similarly as x2 × y3 can'to be disentangled on the grounds that the bases (the x and y)are not the equivalent.
7. Don’t forget to cross check your result.