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Using e to solve equations involving the natural log

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a)

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$\\var{n1}\\ln(\\var{n2}x) = \\var{n3}$

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$x = $[[0]]

\n

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b) $\\var{n4}\\ln(\\frac{\\var{n5}x}{\\var{n6}}) = \\var{n7}$

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$x = $[[1]]

\n

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c) $\\var{n8} = \\ln(\\frac{\\var{n9}}{\\var{n10}x})$

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Give correct to 4 decimal places:

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$x = $[[2]]

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Solve for $x$ in the following, correct to 3 decimal places: 

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(a)

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 $\\var{n1}\\ln(\\var{n2}x) = \\var{n3}$

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$\\ln(\\var{n2}x) = \\frac{\\var{n3}}{\\var{n1}}$

\n

Next,

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$\\var{n2}x = e^{\\frac{\\var{n3}}{\\var{n1}}}$

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$x= \\frac{e^{(\\frac{\\var{n3}}{\\var{n1}})} }{\\var{n2}} = \\var{ans1}$

\n

\n

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(b)

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 $\\var{n4}\\ln(\\frac{\\var{n5}x}{\\var{n6}}) = \\var{n7}$

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$\\ln(\\frac{\\var{n5}x}{\\var{n6}}) =\\frac{ \\var{n7}}{\\var{n4}}$

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Next,

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$\\frac{\\var{n5}x}{\\var{n6}}=e^{\\frac{ \\var{n7}}{\\var{n4}}}$

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$x= e^{(\\frac{\\var{n7}}{\\var{n4}})} \\times \\frac{\\var{n6}}{\\var{n5}} = \\var{ans2}$

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\n

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(c)

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 $\\var{n8} = \\ln(\\frac{\\var{n9}}{\\var{n10}x})$

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$e^{\\var{n8}}=\\frac{\\var{n9}}{\\var{n10}x}$

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$\\var{n10}xe^{\\var{n8}}=\\var{n9}$

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$x= \\frac{\\var{n9}}{(\\var{n10} \\times e^\\var{n8})} = \\var{ans3}$

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