Question

Find the logarithms of \[\dfrac{1}{256}\text{ to base }2\] \[\text{and 0}\text{.3 to base 9}\].

Answer 1

Hint: Write $\dfrac{1}{256}$ in the form of ${{2}^{n}}$ and also change the decimal form of 0.3 into fractional form. Then use properties of logarithms to find the required value. Use base to power conversion formula of logarithms.

Complete step-by-step answer:

First we should understand the term ‘logarithm’ and then we will see the property of logarithm required to solve this question. In mathematics, the logarithm is the inverse function of exponentiation. That means that the logarithm of a given number ‘n’ is the exponent to which another fixed number the base ‘b’ must be raised, to produce that number ‘n’. Common logarithm has base 10, however we can convert it to any number. Let us take an example: consider a number, here I am using 100, so, 100 can be written as 10 raised to the power 2 or mathematically, ${{10}^{2}}$. Now, we have to find the logarithmic value of 100 with 10 as considering the base of the logarithm. In other words, we can interpret the question as ‘to how much must be the power of 10 should be raised, so that it becomes equal to 100’. We know that 10 raised to power 2 is equal to 100, so the answer is 2. Mathematically, it can be written as ${{\log }_{10}}100=2$. Some important formulas for logarithms are:
$\begin{align}
  & {{\log }_{m}}{{n}^{a}}=a{{\log }_{m}}n,\text{ } \\
 & {{\log }_{a}}\left( m\times n \right)={{\log }_{a}}m+{{\log }_{a}}n\text{ } \\
 & \text{lo}{{\text{g}}_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n \\
 & {{\log }_{{{a}^{b}}}}m=\dfrac{1}{b}{{\log }_{a}}m \\
\end{align}$
Now, come to the question, $\dfrac{1}{256}$ can be written as $\dfrac{1}{{{2}^{8}}}={{2}^{-8}}$.
Now, ${{\log }_{2}}{{2}^{-8}}=(-8\times {{\log }_{2}}2)=(-8\times 1)=-8$.
Also, $0.3=\dfrac{3}{10}$. Therefore, ${{\log }_{9}}0.3={{\log }_{9}}\left( \dfrac{3}{10} \right)={{\log }_{{{3}^{2}}}}\left( \dfrac{3}{10} \right)=\dfrac{1}{2}\left( {{\log }_{3}}3-{{\log }_{3}}10 \right)=\dfrac{1}{2}\left( 1-{{\log }_{3}}10 \right)$.

Note: we would not have easily solved the question if we would not have converted $\dfrac{1}{256}$ into exponent form. Log 10 to the base 3 can be calculated by using a calculator. Fractional conversion of 0.3 was necessary.