Question

How do you find the square root of 3 to the ${{4}^{th}}$ power? \[\]

Answer 1

Hint: We recall the definition of exponent or power to base that is $b\times b\times ....\left( n\text{ times} \right)={{b}^{n}}$and square root number $\sqrt{a}={{a}^{\dfrac{1}{2}}}$. We can interpret the phrase as first we find $\sqrt{3}$ is being raised to the ${{4}^{th}}$ power or we can interpret as square root of ${{3}^{4}}$.We evaluate for both the cases. \[\]

Complete answer:
We know that if we multiply a number with itself then we can express the number in exponent form. The number is called base and the number of times we multiply with it is called exponent. If we multiply the base $b$ with itself $n$ number of times then we express the product as
\[b\times b\times b....\left( n\text{ times} \right)={{b}^{n}}\]
Here $n$ is called exponent, power or index of the base. We also know that the square root of number $a$ is a number say $s$ such that
\[s\times s={{s}^{2}}=a\]
We can write the square root as
\[\sqrt{a}={{a}^{\dfrac{1}{2}}}=s\]
We are asked to evaluate the square root of 3 to the ${{4}^{th}}$ power in the question. So we can interpret it as the square root of 3 that is $\sqrt{3}$ raised to the power ${{4}^{th}}$. So we have
\[{{\left( \sqrt{3} \right)}^{4}}={{\left( {{3}^{\dfrac{1}{2}}} \right)}^{4}}\]
We use law of exponent of power raised to a power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ for $a=3,m=\dfrac{1}{2},n=4$ in the above step to have;
\[\begin{align}
  & \Rightarrow {{\left( \sqrt{3} \right)}^{4}}={{3}^{\dfrac{1}{2}\times 4}} \\
 & \Rightarrow {{\left( \sqrt{3} \right)}^{4}}={{3}^{2}} \\
 & \Rightarrow {{\left( \sqrt{3} \right)}^{4}}=3\times 3 \\
 & \Rightarrow {{\left( \sqrt{3} \right)}^{4}}=9 \\
\end{align}\]
We can alternatively interpret the phrase at square root of ${{3}^{4}}$. So we have
\[\sqrt{{{3}^{4}}}={{\left( {{3}^{4}} \right)}^{\dfrac{1}{2}}}\]
We use law of exponent of power raised to a power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ for $a=3,m=4,n=\dfrac{1}{2}$ in the above step to have;
 \[\begin{align}
  & \Rightarrow \sqrt{{{3}^{4}}}={{3}^{4\times \dfrac{1}{2}}} \\
 & \Rightarrow \sqrt{{{3}^{4}}}={{3}^{2}} \\
 & \Rightarrow \sqrt{{{3}^{4}}}=3\times 3 \\
 & \Rightarrow \sqrt{{{3}^{4}}}=9 \\
\end{align}\]

Note: We note that when square root means positive square root unless otherwise mentioned as negative square root. The negative square root of ${{3}^{4}}$ is $-9$. We should remember that if $a=b\ne 0$ then we can raise them to the same power that is ${{a}^{m}}={{b}^{m}}$. Here in the solution ${{s}^{2}}=a$ we have raised both of them to the power $\dfrac{1}{2}$ to have ${{\left( {{s}^{2}} \right)}^{\dfrac{1}{2}}}={{a}^{\dfrac{1}{2}}}\Rightarrow {{s}^{2\times \dfrac{1}{2}}}={{a}^{\dfrac{1}{2}}}\Rightarrow s={{a}^{\dfrac{1}{2}}}$.