// Numbas version: exam_results_page_options {"name": "Indices: Simplification 2a", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Indices: Simplification 2a", "tags": ["category: Indices"], "metadata": {"description": "

Using indices rules to rewrite an expression from $a^\\frac{m}{n}$ to $\\frac{1}{b}$, for integers $a$, $b$, $m$ and $n$.

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Rewrite the following expression as a fraction:

\n

\\[\\var{a^n}^{\\frac{\\var{m}}{\\var{n}}}\\]

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To find $\\var{a^n}^{\\frac{\\var{m}}{\\var{n}}}$, we want to make use of the following 2 rules:

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    \n
  1. $\\left(a^n\\right)^m = a^{n\\times m}$;
  2. \n
  3. $a^{-n} = \\frac{1}{a^n}$.
  4. \n
\n

By rewriting the power $\\frac{\\var{m}}{\\var{n}}$ as a product of $\\var{m} \\times \\frac{1}{\\var{n}}$, we can apply rule 1:

\n

\\[ \\begin{split} \\var{a^n}^{\\frac{\\var{m}}{\\var{n}}} &\\,= \\var{a^n}^{\\left(\\var{m} \\times \\frac{1}{\\var{n}}\\right)} \\\\ &\\,= \\left(\\var{a^n}^\\frac{1}{\\var{n}}\\right)^\\var{m} \\\\ &\\,= \\var{a}^\\var{m}\\end{split} \\]

\n

Then applying rule 2:

\n

\\[ \\begin{split} \\var{a}^\\var{m} &\\,= \\frac{1}{\\var{a}^{\\var{-m}}} \\\\ &\\,= \\frac{1}{\\var{a^(-m)}} \\end{split} \\]

\n

Therefore,

\n

\\[ \\var{a^n}^{\\frac{\\var{m}}{\\var{n}}} =\\frac{1}{\\var{a^(-m)}}. \\]

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