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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). 

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Please complete the following.

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

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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

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For this question, if the answer was $\\left(\\frac{35}{12}\\right)^{11}-24$, then you could enter  (35/12)^(11)-24.

\n

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

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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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