Chapter 6.6, Problem 2BGP

To determine

To Evaluate:

  ${256}^{\frac{3}{8}}$

Answer to Problem 2BGP

8

Explanation of Solution

Concept Used :

• Definition of nth root:
• Words : For any real numbers a and b , and any positive integer n , if $a^n=b$ then a is the nth root of b . That is $a=\sqrt[n]{b}$

    $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$

• For any real number b and any positive inteber n , $b^{\frac{1}{n}}=\sqrt[n]{b}$ , except when b < 0 and n is even. When b < 0 and n is even , a complex root may exist.
• $\sqrt{-1}=i$
• $b^{-n}=\frac{1}{b^n}$

Calculation:

• Given${256}^{\frac{3}{8}}$
• $\begin{array}{l}{256}^{\frac{3}{8}}\\ =\sqrt[8]{{256}^3}\mathrm{...............}[\text{definition of }{a}^{\frac{1}{n}}]\\ =\sqrt[8]{{\left(2^8\right)}^3}\\ =\sqrt[8]{{\left(2^3\right)}^8}\mathrm{......................}[{\left(a^x\right)}^y=a^{x\cdot y}=a^{y\cdot x}={\left(a^y\right)}^x]\\ =2^3\mathrm{..............................}[\sqrt[n]{b^n}=b]\\ =8\end{array}$
• So, ${256}^{\frac{3}{8}}=8$