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Question: How do you write \( {2^{ - 3}} = \dfrac{1}{8} \) in logarithmic form?...

How do you write 2−3=18{2^{ - 3}} = \dfrac{1}{8} in logarithmic form?

Explanation

Solution

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⁡bx=y{\log _b}x = y is when converted into exponential form is equivalent to by=x{b^y} = x ,so compare with this form and form your answer accordingly.

Complete step-by-step answer :
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
Any logarithmic form log⁡bx=y{\log _b}x = y when converted into equivalent exponential form results in by=x{b^y} = x
So in Our question we are given 2−3=18{2^{ - 3}} = \dfrac{1}{8} and if compare this with log⁡bx=y{\log _b}x = y we get
b=2 y=−3 x=18   b = 2 \\\ y = - 3 \\\ x = \dfrac{1}{8} \;
Hence the logarithmic form of 2−3=18{2^{ - 3}} = \dfrac{1}{8} will be equivalent to log⁡218=−3{\log _2}\dfrac{1}{8} = - 3 .
Therefore, our desired answer is log⁡218=−3{\log _2}\dfrac{1}{8} = - 3 .
So, the correct answer is “ log⁡218=−3{\log _2}\dfrac{1}{8} = - 3 .”.

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}