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((bc-bb)/bxcoeff)^(1/bpower)

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intsoln^intpower

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If  $\\sqrt[\\var{intpower}]{x}=\\var{intrhs}$, then $x=$ [[0]].

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If  $\\simplify{{bxcoeff}y^(1/{bpower})+{bb}}=\\var{bc}$, then $y=$ [[0]].

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For this question, if the answer was $\\left(\\frac{35}{11}\\right)^{11}-24$, then you could enter  (35/11)^(11)-24.

\n

If  $\\displaystyle{\\simplify{(root(z+{db},{dpower}))/{ddenom}}}=\\var{dc}$, then $z=$ [[0]].

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a) Given $\\sqrt[\\var{intpower}]{x}=\\var{intrhs}$, we raise both sides to the power of $\\var{intpower}$ to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $\\sqrt[\\var{intpower}]{x}$ $=$ $\\var{intrhs}$ $\\left(\\sqrt[\\var{intpower}]{x}\\right)^{\\var{intpower}}$ $=$ $\\simplify[basic]{({intrhs})^{intpower}}$ $x$ $=$ $\\var{intsoln}$
\n

\n

b) Given $\\simplify{{bxcoeff}y^(1/{bpower})+{bb}}=\\var{bc}$, we can rearrange the equation to get $y^\\frac{1}{\\var{bpower}}$ by itself and then we can raise both sides to the power of $\\var{bpower}$ to get $y$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $\\simplify{{bxcoeff}y^(1/{bpower})+{bb}}$ $=$ $\\var{bc}$ $\\simplify{{bxcoeff}y^(1/{bpower})}$ $=$ $\\simplify[basic]{{bc}-{bb}}$ $\\simplify{{bxcoeff}y^(1/{bpower})}$ $=$ $\\simplify{{bc-bb}}$ $y^\\frac{1}{\\var{bpower}}$ $=$ $\\simplify[!basic]{{bc-bb}/{bxcoeff}}$ $y^\\frac{1}{\\var{bpower}}$ $=$ $\\simplify{{bc-bb}/{bxcoeff}}$ $\\left(y^\\frac{1}{\\var{bpower}}\\right)^{\\var{bpower}}$ $=$ $\\simplify[basic]{({(bc-bb)/bxcoeff})^{bpower}}$ $y$ $=$ $\\var{bsoln}$
\n

\n

c) Given $\\displaystyle{\\simplify{(root(z+{db},{dpower}))/{ddenom}}}=\\var{dc}$, we can rearrange the equation to get $\\simplify{(root(z+{db},{dpower}))}$ by itself, then we can raise both sides to the power of $\\var{dpower}$, and finally rearrange to get $z$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $\\displaystyle{\\simplify{(root(z+{db},{dpower}))/{ddenom}}}$ $=$ $\\var{dc}$ $\\displaystyle{\\simplify{(root(z+{db},{dpower}))}}$ $=$ $\\simplify[basic]{{dc}*{ddenom}}$ $\\displaystyle{\\simplify{(root(z+{db},{dpower}))}}$ $=$ $\\var{dc*ddenom}$ $\\left(\\sqrt[\\var{dpower}]{\\simplify{z+{db}}}\\right)^\\var{dpower}$ $=$ $\\simplify[basic]{({dc*ddenom})^{dpower}}$ $\\simplify{z+{db}}$ $=$ $\\simplify[basic]{-{abs(dc*ddenom)}^{dpower}}$  $\\simplify[basic]{({abs(dc*ddenom)})^{dpower}}$  $\\simplify[basic]{({(dc*ddenom)})^{dpower}}$ $z$ $=$ $\\simplify[basic]{-{abs(dc*ddenom)}^{dpower}-{db}}$  $\\simplify[basic]{({abs(dc*ddenom)})^{dpower}-{db}}$  $\\simplify[basic]{({(dc*ddenom)})^{dpower}-{db}}$
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Questions to test if the student knows the inverse of fractional power or root (and how to solve equations that contain them).

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