Question

What is ${{2}^{\dfrac{4}{3}}}$ equal to?

Answer 1

Hint: We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ . We must use the converse of this identity to get the equation in the form ${{\left( {{a}^{m}} \right)}^{n}}$ . After evaluating ${{a}^{m}}$ , we must sensibly split it into factors remembering the identities ${{\left( b\times c \right)}^{x}}={{b}^{x}}\times {{c}^{x}}$ and ${{p}^{\dfrac{1}{x}}}=\sqrt[x]{p}$ . Using these identities, we can convert ${{2}^{\dfrac{4}{3}}}$ into the required form.

Complete step by step solution:
We can easily write
${{2}^{\dfrac{4}{3}}}={{2}^{4\times \dfrac{1}{3}}}...\left( i \right)$
We know by the laws of exponent that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ , where a, m and n are rational numbers.
Using the converse of this identity on the RHS part of equation (i), we get
${{2}^{\dfrac{4}{3}}}={{\left( {{2}^{4}} \right)}^{\dfrac{1}{3}}}$
Now, we will evaluate the ${{2}^{4}}$ part of the RHS. Our equation thus becomes,
${{2}^{\dfrac{4}{3}}}={{\left( 2\times 2\times 2\times 2 \right)}^{\dfrac{1}{3}}}$
\[\Rightarrow {{2}^{\dfrac{4}{3}}}={{\left( 16 \right)}^{\dfrac{1}{3}}}...\left( ii \right)\]
Here, we should note that 16 could be split into factors of 8 and 2. The reason we are splitting 16 into these two factors is that 8 is a perfect cube and the remaining part, i.e., 2 is a prime number. This will be clearer in the following steps.
We can rewrite equation (ii) by splitting 16 into 8 and 2.
\[{{2}^{\dfrac{4}{3}}}={{\left( 8\times 2 \right)}^{\dfrac{1}{3}}}...\left( iii \right)\]
We know about the property ${{\left( a\times b \right)}^{x}}={{a}^{x}}\times {{b}^{x}}$ . Using this property in equation (iii), we get
\[{{2}^{\dfrac{4}{3}}}={{\left( 8 \right)}^{\dfrac{1}{3}}}\times {{\left( 2 \right)}^{\dfrac{1}{3}}}...\left( iv \right)\]
We must remember the basic identity of exponentials ${{a}^{\dfrac{1}{x}}}=\sqrt[x]{a}$ , to further simplify the RHS of equation (iv),
\[{{2}^{\dfrac{4}{3}}}=\sqrt[3]{8}\times \sqrt[3]{2}...\left( v \right)\]
It is pretty clear that the cube root of 8, i.e., $\sqrt[3]{8}$ is equal to 2. To attain this level of simplicity, we took 8 as one of the factors.
Thus, substituting the value of $\sqrt[3]{8}$ in equation (v), we get the following equation,
\[{{2}^{\dfrac{4}{3}}}=2\times \sqrt[3]{2}\]
\[\Rightarrow {{2}^{\dfrac{4}{3}}}=2\sqrt[3]{2}\]
Hence, the required value of ${{2}^{\dfrac{4}{3}}}$ is \[2\sqrt[3]{2}\] .

Note: Alternatively, a better approach could be to split $\dfrac{4}{3}$ into $\left( 1+\dfrac{1}{3} \right)$ . Using the identity ${{\left( a \right)}^{m+n}}={{\left( a \right)}^{m}}\times {{\left( a \right)}^{n}}$ , we could write
\[{{2}^{\dfrac{4}{3}}}={{\left( 2 \right)}^{1+\dfrac{1}{3}}}\]
\[\Rightarrow {{2}^{\dfrac{4}{3}}}={{\left( 2 \right)}^{1}}\times {{\left( 2 \right)}^{\dfrac{1}{3}}}\]
Now, if we use the identity ${{a}^{\dfrac{1}{x}}}=\sqrt[x]{a}$ , we get
\[{{2}^{\dfrac{4}{3}}}=2\sqrt[3]{2}\] .