Question

How do you express the number in standard notation $2.42\times {{10}^{-4}}$?

Answer 1

Hint: First convert the decimal to fraction by dividing ${{10}^{2}}$. Then write ${{10}^{-4}}$ as $\dfrac{1}{{{10}^{4}}}$ . Multiply the terms in the denominator by adding the powers of ‘10’. After that decide the standard notation according to the zeros or according to the decimal places.

Complete step-by-step solution:
As we know, if there is ‘n’ number of digits after the decimal, then the denominator of fraction will be ${{10}^{n}}$
So, $2.42\times {{10}^{-4}}$ can be written in fraction as
$=\dfrac{242}{100}\times {{10}^{-4}}$
‘100’ can be written as ${{10}^{2}}$ and ${{10}^{-4}}$ can be written as $\dfrac{1}{{{10}^{4}}}$.
So, $\dfrac{242}{100}\times {{10}^{-4}}$ becomes
$\begin{align}
  & =\dfrac{242}{{{10}^{2}}}\times \dfrac{1}{{{10}^{4}}} \\
 & =\dfrac{242}{{{10}^{2}}\times {{10}^{4}}} \\
\end{align}$
Again, as we know
${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$
So, $\dfrac{242}{{{10}^{2}}\times {{10}^{4}}}=\dfrac{242}{{{10}^{2+4}}}=\dfrac{242}{{{10}^{6}}}$
Standard notation: Standard notation is the normal way of writing numbers.
So, $\dfrac{242}{{{10}^{6}}}$ can be written in standard notation as
$=0.000242$
(for ${{10}^{6}}$ there will be ‘6’ digits after the decimal, so ‘3’ zeros are adjusted since there are only ‘3’ digits present in the numerator. )
Hence, the standard notation of $2.42\times {{10}^{-4}}=0.000242$. This is the required solution.

Note: We can directly write in standard notation by moving the decimal only. The number of decimal places moved will be the exponent on 10. If the exponent is positive then we have moved the decimal to the right and if the exponent is negative then we have moved the decimal to left. For example, for the standard notation of $2.42\times {{10}^{-4}}$, we have to move 4 decimal places to the left. So, the ‘3’ zeros will fill the places to give the standard notation as $0.000242$.