Question

To calculate: The single radical of the expression \(\displaystyle\sqrt{{{x}\sqrt{{{x}\sqrt{{{x}}}}}}}\)

Answer 1

Step 1
Consider the expression \(\displaystyle\sqrt{{{x}\sqrt{{{x}\sqrt{{{x}}}}}}}\)
\(\displaystyle\sqrt{{{x}\sqrt{{{x}\sqrt{{{x}}}}}}}={\left({x}{\left({x}{\left({x}^{{\frac{{1}}{{2}}}}\right)}\right)}^{{\frac{{1}}{{2}}}}\right)}^{{\frac{{1}}{{2}}}}\)
Use the product property and the expression becomes,
\(\displaystyle{\left({x}{\left({x}{\left({x}^{{\frac{{1}}{{2}}}}\right)}\right)}^{{\frac{{1}}{{2}}}}\right)}^{{\frac{{1}}{{2}}}}={\left({x}{\left({x}^{{{1}+\frac{{1}}{{2}}}}\right)}^{{\frac{{1}}{{2}}}}\right)}^{{\frac{{1}}{{2}}}}\)
\(\displaystyle={\left({x}{\left({x}^{{\frac{{3}}{{2}}}}\right)}^{{\frac{{1}}{{2}}}}\right)}^{{\frac{{1}}{{2}}}}\)
\(\displaystyle{\left({x}\cdot\ {x}^{{\frac{{3}}{{4}}}}\right)}^{{\frac{{1}}{{2}}}}\)
Add the powers of the same bases.
\(\displaystyle{\left({x}\cdot\ {x}^{{\frac{{3}}{{4}}}}\right)}^{{\frac{{1}}{{2}}}}={\left({x}^{{{1}+\frac{{3}}{{4}}}}\right)}^{{\frac{{1}}{{2}}}}\)
\(\displaystyle={\left({x}^{{\frac{{7}}{{4}}}}\right)}^{{\frac{{1}}{{2}}}}\)
\(\displaystyle={x}^{{\frac{{7}}{{8}}}}\)
The obtained expression with rational exponents can be rewritten into radicals as.
\(\displaystyle{x}^{{\frac{{7}}{{8}}}}={\left({x}^{{{7}}}\right)}^{{\frac{{1}}{{8}}}}\)
\(=\sqrt[8]{x^{7}}\)
Therefore, the value of the expression is \(=\sqrt[8]{x^{7}}\)

Answer 2

Answer is given below (on video)