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Multiply two numbers in standard form, then divide two numbers in standard form.
\nNeeds marking algorithm to allow equal values in standard form to gain equal marks
"}, "tags": ["L0", "SN"], "extensions": [], "preamble": {"js": "", "css": ""}, "variable_groups": [], "statement": "Simplify, leaving the result in standard index form.
\nWrite
We use the rules for multiplying and dividing powers.
\nTo multiply powers we add, for example $10^2 \\times 10^6 = 10^{(2+6)} = 10^{8}$.
\nTo divide powers we subtract, for example $10^2 \\div 10^6 = 10^{(2-6)} = 10^{-4}$.
\n\n
\\[ \\begin{align} (\\var{int} \\times 10^\\var{ran - 3} ) \\times (\\var{int - 11} \\times 10^\\var{ran}) &= \\var{int} \\times (\\var{int-11}) \\times 10^{(\\var{ran - 3} + \\var{ran})} \\\\&= (\\var{int*(int-11)}) \\times 10^{\\var{ran - 3 + ran}} \\end{align} \\]
\nWe want to write this number in standard form so we move decimal place one place to the left to get
\n\\[(\\var{int*(int-11)/10}) \\times 10^{(\\var{ran - 3 + ran} + 1)} = (\\var{int*(int-11)/10}) \\times 10^{\\var{ran - 2 + ran}} \\text{.}\\]
\nSimilarly,
\n\\[ \\begin{align} (\\var{int*int} \\times 10^\\var{ran -4} ) \\div (\\var{int} \\times 10^\\var{ran - 2}) &= \\var{int*int} \\div \\var{int} \\times 10^{(\\var{ran - 4} - \\var{ran -2})} \\\\&= \\var{int} \\times 10^{-2} \\text{.}\\end{align} \\]
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