Question

 Consider the value ${{\log }_{10}}2=0.3010$ ,then the value of ${{\log }_{2}}10$ is equal to
(a) $\dfrac{699}{301}$
(b) $\dfrac{1000}{301}$
(c) 0.3010
(d) 0.6990

Answer 1

Hint: At first use the identity, ${{\log }_{c}}d=\dfrac{\log d}{\log c}$ and take multiplicative inverse and write it as $\dfrac{\log c}{\log d}=\dfrac{1}{{{\log }_{c}}d}$, then substitute ‘d’ as 2 and ‘c’ as 10.

Complete step-by-step answer:
In the question the value of ${{\log }_{10}}2$ is given as 0.3010 and we have to find the value of ${{\log }_{2}}10$ .
So, we will first know what logarithms.
In mathematics, the logarithm is the inverse function to exponentiation. That means the logarithm of a given number $x$ is the exponent to which another fixed number, the base b must be raised, to produce that number $x$ .In the simplest case, the logarithm counts the number of the same factor in repeated multiplication for example since 1000 = 10 x 10 x 10 which is also written as ${{10}^{3}}$ so in logarithm we can write it as ${{\log }_{10}}1000=3$ .The logarithm of $x$ to base b is written as ${{\log }_{b}}x$.
More generally exponentiation allows only real number as base to be raised to real power, always producing a positive result, So ${{\log }_{b}}x$ for any two positive real numbers b and \[x\] , where b is not equal to 1, is always a unique real number y. More explicitly the defining relation between exponential and logarithm is \[{{\log }_{b}}x\] exactly if ${{b}^{y}}=x$ and $x>0$ , $b>0$ and $b\ne 1$ .
The logarithm functions have some identities such as,
\[\begin{align}
  & {{\log }_{b}}x+{{\log }_{b}}y={{\log }_{b}}xy \\
 & {{\log }_{b}}x-{{\log }_{b}}y={{\log }_{b}}\dfrac{x}{y} \\
 & c{{\log }_{b}}x={{\log }_{b}}{{x}^{c}} \\
 & {{\log }_{a}}b=\dfrac{1}{{{\log }_{b}}a}\text{ and many others} \\
 & \\
\end{align}\]
In the question we are given the value of \[\text{lo}{{\text{g}}_{10}}2=0.3010\] .
So value of ${{\log }_{2}}10$ will be $\dfrac{1}{{{\log }_{10}}2}$ as ${{\log }_{a}}b=\dfrac{1}{{{\log }_{b}}a}$
So ${{\log }_{2}}10=\dfrac{1}{0.3010}=\dfrac{1000}{301}$
Hence, the correct option is (b).

Note: Students should study about logarithms and should have knowledge about the identities related to it. Students should know how to change or inverse the given logarithm terms by exchanging bases because this is the concept generally used in these types of questions.