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Question: How do you evaluate \[{{\log }_{2}}\left( \dfrac{1}{16} \right)\]?...

How do you evaluate log⁡2(116){{\log }_{2}}\left( \dfrac{1}{16} \right)?

Explanation

Solution

From the question given, we have been asked to evaluate log⁡2(116){{\log }_{2}}\left( \dfrac{1}{16} \right). We can clearly observe that the given expression is in logarithmic form. We can solve the given logarithmic expression by using some basic logarithmic formulae. First of all, let us assume the given logarithmic expression as zz.

Complete step-by-step solution:
Now from the question we have log⁡2(116){{\log }_{2}}\left( \dfrac{1}{16} \right)
By assuming the given logarithmic expression as zz, we can write it as z=log⁡2(116)z={{\log }_{2}}\left( \dfrac{1}{16} \right)
Now, as we have already discussed above, by using some basic laws of logarithms, we can evaluate it.
From the basic laws of logarithms, we can write the above equation as z=−log⁡2(16)z=-{{\log }_{2}}\left( 16 \right)
Shift the negative sign from the right hand side of the equation to the left hand side of the equation.
−z=log⁡2(16)-z={{\log }_{2}}\left( 16 \right)
Now, we know that 1616 can be written as 24{{2}^{4}}.
Now, by doing this, we get the equation as,
−z=log⁡2(24)-z={{\log }_{2}}\left( {{2}^{4}} \right)
On furthermore simplification, we get
−z=4log⁡2(2)-z=4{{\log }_{2}}\left( 2 \right)
In logarithms, we have one basic formula log⁡aa=1{{\log }_{a}}a=1.
By using the above basic formula of logarithms, we get the equation as,
−z=4(1)-z=4\left( 1 \right)
⇒−z=4\Rightarrow -z=4
⇒z=−4\Rightarrow z=-4
Hence, the given logarithmic expression is evaluated.
As we have been already discussed above, by using some basic formulae of logarithms, we get the given logarithmic expression evaluated.

Note: We should be well aware of the logarithms. Also, we should be well known about the basic laws and formulae of logarithms concept. We should also well know about the application of each and every formula of logarithms. Also, we should be very careful while doing the calculation of the logarithmic equation, as it was a little confusing to solve. This can be simply written as log⁡2(116)=log⁡2(124)=log⁡2(2−4)=−4{{\log }_{2}}\left( \dfrac{1}{16} \right)={{\log }_{2}}\left( \dfrac{1}{{{2}^{4}}} \right)={{\log }_{2}}\left( {{2}^{-4}} \right)=-4