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} $