Question

express $17^{\frac{11}{6}}$ in radical form.
$= 17\sqrt6{1419857}$
$= 17\sqrt6{1419857}$
$= 17\sqrt6{1419857^1}$

Explanation:

Step1: Recall the exponent - radical rule

The rule for converting an expression of the form \(a^{\frac{m}{n}}\) to radical form is \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\) (or equivalently \(a^{\frac{m}{n}} = a\sqrt[n]{a^{m - n}}\) when \(m>n\) and we can simplify the coefficient). First, let's recall the general formula for converting a rational exponent to a radical: for any real number \(a\) and positive integers \(m,n\) with \(n>1\), \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\).

Step2: Apply the rule to \(17^{\frac{11}{6}}\)

We can rewrite the exponent \(\frac{11}{6}\) as \(1+\frac{5}{6}\), so \(17^{\frac{11}{6}}=17^{1 + \frac{5}{6}}\). Using the property of exponents \(a^{x + y}=a^{x}\times a^{y}\), we get \(17^{1+\frac{5}{6}}=17^{1}\times17^{\frac{5}{6}}\). Now, convert \(17^{\frac{5}{6}}\) to radical form. According to the rule \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\), when \(a = 17\), \(m = 5\) and \(n=6\), we have \(17^{\frac{5}{6}}=\sqrt[6]{17^{5}}\). But we can also calculate \(17^{5}\): \(17^{1}=17\), \(17^{2}=289\), \(17^{3}=17\times289 = 4913\), \(17^{4}=17\times4913=83521\), \(17^{5}=17\times83521 = 1419857\). So \(17^{\frac{5}{6}}=\sqrt[6]{17^{5}}=\sqrt[6]{1419857}\). Then \(17^{\frac{11}{6}}=17\times\sqrt[6]{1419857}\), which is written as \(17\sqrt[6]{1419857}\) (the extra "1" in the last box of the given image is a typo, the correct form is without the "1" inside the radical or as a coefficient).

Answer:

\(17\sqrt[6]{1419857}\)