Question

How do you write ${\left( {\dfrac{1}{4}} \right)^{ - 6}} = 4096$ in logarithmic form?

Answer 1

Hint : In order to determine the value of the above question in logarithmic form ,use the definition of logarithm that the logarithm of the form ${\log _b}x = y$ is when converted into exponential form is equivalent to ${b^y} = x$ ,so compare with this form and form your answer accordingly.

Complete step-by-step answer :
We are given ${\left( {\dfrac{1}{4}} \right)^{ - 6}} = 4096$
Removing negative sign of exponent by taking the reciprocal and writing $4 = {2^2}$
${\left( {{2^2}} \right)^6} = 4096 \\\ {\left( 2 \right)^{12}} = 4096 \;$
To convert the above into logarithmic form, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
Any logarithmic form ${\log _b}x = y$ when converted into equivalent exponential form results in ${b^y} = x$
So in Our question we are given ${\left( 2 \right)^{12}} = 4096$ and if compare this with ${\log _b}x = y$ we get
$b = 2 \\\ y = 12 \\\ x = 4096 \;$
Hence the logarithmic form of ${\left( 2 \right)^{12}} = 4096$ will be equivalent to ${\log _2}4096 = 12$ .
Therefore, our required answer is ${\log _2}4096 = 12$ .
So, the correct answer is “ ${\log _2}4096 = 12$ ”.

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 .
${\log _b}(mn) = {\log _b}(m) + {\log _b}(n)$
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
${\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)$
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
$n\log m = \log {m^n}$