This question aims to test understanding and ability to use the laws of indices.
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\n$a^{\\var{x}} : a^{\\var{y}} =$ [[1]].(hier mag je de eventuele negatieve exponent laten staan)
\n$(a^{\\var{x}})^{\\var{y}} =$ [[2]].
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\n$a^{m} \\cdot a^n = a^{m+n}$
\n$a^{m} : a^n = a^{m-n}$.
\n$(a^{m})^n = a^{m \\cdot n}$.
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\n$\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q} =$ [[0]].
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\n$\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})} =$ [[0]].
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\n$a^m \\div a^n = a^{m-n}$.
\n$a^{-m} = \\displaystyle\\frac{1}{a^m}$.
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\n$(\\simplify{{c}*a^{p}})^{\\var{q}} =$ [[0]].
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\n$(ab)^m = a^mb^m$.
\n$(a^m)^n = a^{mn}$.
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\nUsing this rule,
\n\\[
\\begin{align}
a^\\var{x} \\times a^\\var{y}\\ &= a^\\simplify[all, !collectNumbers]{{x}+{y}}\\\\
&= a^\\var{x+y}.
\\end{align}
\\]
We are asked to find $\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}$.
\nNotice there is a constant in front of each of the terms.
\nTo do this, write the product out explicitly, as
\n\\[\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q} = \\var{c} \\times \\var{d} \\times a^\\var{p} \\times a^\\var{q}.\\]
\nWe know that $\\var{c} \\times \\var{d} = \\var{c*d}$, and using the rule of indices: $a^\\var{p} \\times a^\\var{q} = a^\\var{p+q}$.
\nTherefore:
\n\\begin{align}
\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}&= \\var{c*d} \\times a^\\var{p+q} \\\\
&= \\simplify{{c*d}*a^{p+q}}.
\\end{align}
Here we are using: $a^m \\div a^n = a^{m-n}$.
\nWe are asked to simplify the expression, $\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})}$.
\nTo do this, we just have to use the previously mentioned rule of indices. We write this out explicity as
\n\\[\\simplify{{b}*a^{x}/({g}*a^{y})} = \\simplify{{b}/{g}} \\times \\simplify{a^{x}/(a^{y})}.\\]
\nUsing rules of indices,
\n\\begin{align} \\frac{a^\\var{x}}{a^\\var{y}} &= a^\\var{x} \\div a^\\var{y}\\\\
&= a^\\simplify[all, !collectNumbers]{{x}-{y}}\\\\
&= a^\\var{x-y}.
\\end{align}
Therefore,
\n\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\simplify{a^{{x}-{y}}}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}
Alternatively,
\nUsing the rule of indices: $a^{-m} = \\displaystyle\\frac{1}{a^{m}}$, we can rewrite the question as:
\n\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\frac{a^\\var{x}}{a^\\var{y}}\\\\
&= \\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}}.
\\end{align}
And then using the rule: $a^m \\times a^n = a^{m+n}$, this becomes:
\n\\begin{align}
\\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}} &= \\simplify{{b}/{g}} \\times a^\\simplify[all,!collectNumbers]{{x}+(-{y})}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}
The question asks us to simplify $(\\simplify{{c}*a^{p}})^{\\var{q}}$.
\nTo do this we use the rules:
\n\\[(a^{m})^{n} = a^{mn},\\]
\n\\[(ab)^m = a^mb^m.\\]
\nWe can then expand the equation as
\n\\[(\\simplify{{c}*a^{p}})^{\\var{q}}= \\var{c}^{\\var{q}} \\times (a^{\\var{p}})^{\\var{q}}.\\]
\nThen using the rule of indices mentioned previously,
\n\\[
\\begin{align}
(\\simplify{{c}*a^{p}})^{\\var{q}}&= \\simplify{{c}^{q}} \\times a^\\var{p*q}\\\\
&= \\simplify{{c}^{q}*a^{p*q}}.
\\end{align}
\\]
The question asks us to simplify $\\sqrt[\\var{d}]{\\var{x}^\\var{d}a}$.
\nTo do this we use the rules:
\n\\[a^\\frac{1}{m} = \\sqrt[m]{a},\\]
\n\\[(ab)^m = a^mb^m.\\]
\nWe can expand the expression as follows:
\n\\[
\\begin{align}
\\sqrt[\\var{d}]{a} &= (\\simplify{a})^\\frac{1}{\\var{d}}\\\\
&= a^\\frac{1}{\\var{d}}.
\\end{align}
\\]
The question requires us to simplify $\\sqrt[\\var{c}]{a^\\var{q}}$.
\nHere, we use the rule of indices: $a^\\frac{n}{m} = \\sqrt[m]{a^n}$, allowing us to expand the expression as follows:
\n\\[
\\begin{align}
\\sqrt[\\var{c}]{\\simplify{a^{q}}} &= \\simplify[fractionnumbers,all]{(a^{q})^{{1}/{{c}}}}\\\\
&= \\simplify[fractionnumbers,all]{a^{{q}/{c}}}.
\\end{align}
\\]