How do you simplify \[{(2)^{\dfrac{3}{6}}}\]?
Answer
622.8k+ views
Hint: Here, we will first check if the exponent of the given term can be simplified further or not. If it can be simplified, then we will reduce the exponent to the simplest form. Then we will use the radical formula to simplify the equation and find the required answer.
Formula used:
We will use the radical formula \[{a^{\dfrac{m}{n}}} = \sqrt[n]{{{a^m}}}\].
Complete step by step solution:
We are required to simplify the expression \[{(2)^{\dfrac{3}{6}}}\].
Let us first check if the exponent can be simplified further or not.
The exponent given is \[\dfrac{3}{6}\].
We see that the numerator and the denominator have a common factor, which is 3.
So, we can cancel out the common factor 3 from both the numerator and the denominator. This gives us
\[\dfrac{3}{6} = \dfrac{{3 \times 1}}{{3 \times 2}} = \dfrac{1}{2}\]
Now, we see that in \[\dfrac{1}{2}\], both the numerator and denominator have only the common factor 1. Thus, the exponent has reduced to the simplest form.
We shall now raise the base, which is 2 to the exponent \[\dfrac{1}{2}\]. Therefore, we get
\[{(2)^{\dfrac{3}{6}}} = {(2)^{\dfrac{1}{2}}}\]
Using the law \[{a^{\dfrac{m}{n}}} = \sqrt[n]{{{a^m}}}\], where \[a = 2\], \[m = 1\] and \[n = 2\], we get
\[ \Rightarrow {(2)^{\dfrac{3}{6}}} = \sqrt[2]{{{2^1}}}\]
This means that \[{(2)^{\dfrac{1}{2}}}\] is the same as the square root of 2. Since \[\sqrt 2 \] is an irrational number, we cannot simplify this further. Hence,
\[ \Rightarrow {(2)^{\dfrac{3}{6}}} = \sqrt 2 \]
Therefore \[{(2)^{\dfrac{3}{6}}} = \sqrt 2 \].
Note:
Another way to approach this problem is by the application of logarithms.
Let us take \[x = {(2)^{\dfrac{3}{6}}}\]
Simplify the exponent, we get
\[ \Rightarrow x = {(2)^{\dfrac{1}{2}}}\]
Now, applying logarithms on both sides, we get
\[ \Rightarrow \log x = \log \left( {{2^{\dfrac{1}{2}}}} \right)\]
Using the property \[\log {a^m} = m\log a\], we have
\[ \Rightarrow \log x = \dfrac{1}{2}\log 2\]
By using logarithm tables, we see that \[\log 2 = 0.693\] (approximately). So,
\[ \Rightarrow \log x = \dfrac{1}{2} \times 0.693\]
Dividing the expression by 2, we have
\[ \Rightarrow \log x = 0.3465\]
To find \[x\], we will take an exponential function on both sides of the above equation. Therefore, we get
\[ \Rightarrow x = {e^{0.3465}} = 1.414\] (approximately)
Now, we know that \[\sqrt 2 \approx 1.414\], which is the same value as above.
Therefore, \[{(2)^{\dfrac{3}{6}}} = \sqrt 2 \].
Formula used:
We will use the radical formula \[{a^{\dfrac{m}{n}}} = \sqrt[n]{{{a^m}}}\].
Complete step by step solution:
We are required to simplify the expression \[{(2)^{\dfrac{3}{6}}}\].
Let us first check if the exponent can be simplified further or not.
The exponent given is \[\dfrac{3}{6}\].
We see that the numerator and the denominator have a common factor, which is 3.
So, we can cancel out the common factor 3 from both the numerator and the denominator. This gives us
\[\dfrac{3}{6} = \dfrac{{3 \times 1}}{{3 \times 2}} = \dfrac{1}{2}\]
Now, we see that in \[\dfrac{1}{2}\], both the numerator and denominator have only the common factor 1. Thus, the exponent has reduced to the simplest form.
We shall now raise the base, which is 2 to the exponent \[\dfrac{1}{2}\]. Therefore, we get
\[{(2)^{\dfrac{3}{6}}} = {(2)^{\dfrac{1}{2}}}\]
Using the law \[{a^{\dfrac{m}{n}}} = \sqrt[n]{{{a^m}}}\], where \[a = 2\], \[m = 1\] and \[n = 2\], we get
\[ \Rightarrow {(2)^{\dfrac{3}{6}}} = \sqrt[2]{{{2^1}}}\]
This means that \[{(2)^{\dfrac{1}{2}}}\] is the same as the square root of 2. Since \[\sqrt 2 \] is an irrational number, we cannot simplify this further. Hence,
\[ \Rightarrow {(2)^{\dfrac{3}{6}}} = \sqrt 2 \]
Therefore \[{(2)^{\dfrac{3}{6}}} = \sqrt 2 \].
Note:
Another way to approach this problem is by the application of logarithms.
Let us take \[x = {(2)^{\dfrac{3}{6}}}\]
Simplify the exponent, we get
\[ \Rightarrow x = {(2)^{\dfrac{1}{2}}}\]
Now, applying logarithms on both sides, we get
\[ \Rightarrow \log x = \log \left( {{2^{\dfrac{1}{2}}}} \right)\]
Using the property \[\log {a^m} = m\log a\], we have
\[ \Rightarrow \log x = \dfrac{1}{2}\log 2\]
By using logarithm tables, we see that \[\log 2 = 0.693\] (approximately). So,
\[ \Rightarrow \log x = \dfrac{1}{2} \times 0.693\]
Dividing the expression by 2, we have
\[ \Rightarrow \log x = 0.3465\]
To find \[x\], we will take an exponential function on both sides of the above equation. Therefore, we get
\[ \Rightarrow x = {e^{0.3465}} = 1.414\] (approximately)
Now, we know that \[\sqrt 2 \approx 1.414\], which is the same value as above.
Therefore, \[{(2)^{\dfrac{3}{6}}} = \sqrt 2 \].
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