// Numbas version: finer_feedback_settings {"name": "Logaritme- og eksponentiallikninger", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Logaritme- og eksponentiallikninger", "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "

Løs likningene. Avrund svaret til 2 desimaler.

", "advice": "

Løsningsformelen for en logaritmelikning på grunnform ser slik ut:

$\\begin{align}\\ln(x)&=a\\\\ &\\Downarrow \\\\x&=e^{a}\\end{align}$

Vi snur på likningen og regner ut:

$\\begin{align}
\\simplify{{a1} ln(x)+{b1}}&=\\var{c1}\\\\
\\simplify{{a1}ln(x)} &= \\simplify[!collectNumbers]{{c1}-{b1}}\\\\
\\simplify{{a1}ln(x)}&= \\simplify{{c1}-{b1}}\\\\
\\ln(x) &= \\simplify{{{c1}-{b1}}/{{a1}}}\\\\
x &= e^{\\simplify{{{c1}-{b1}}/{{a1}}}}\\\\
\\end{align}$

$x=${precround({exp(({c1}-{b1})/{a1})},2)}

Likningen $\\simplify{{a2}(ln(x))^{2}+{b2}ln(x)+{c2}=0}$ er en andregradslikning med $\\ln(x)$ som ukjent. Vi bruker da først abc-formelen for å finne $\\ln(x)$:

$a=\\var{a2},\\;b=\\var{b2},\\;c=\\var{c2}$

$\\ln(x)=\\dfrac{\\simplify{-{b2}}\\pm\\sqrt{\\simplify[!collectNumbers,timesDot]{{b2}^{2}-4*{a2}*{c2}}}}{2\\cdot\\var{a2}}=\\dfrac{\\simplify{-{b2}}\\pm\\sqrt{\\simplify{{b2}^2-4*{a2}*{c2}}}}{\\simplify{2*{a2}}}$.

Vi to løsninger for $\\ln(x)$: $\\ln(x)= \\var{x[0]}\\text{ og } \\ln(x)=\\var{x[1]}$.

Vi løser for $x$:

$\\ln(x)= \\var{x[0]}\\Rightarrow x=e^{\\var{x[0]}}=${precround(e^(x[0]),6)}$\\approx\\underline{\\underline{\\var{sol[0]}}}$

$\\ln(x)= \\var{x[1]}\\Rightarrow x=e^{\\var{x[1]}}=${precround(e^(x[1]),6)}$\\approx\\underline{\\underline{\\var{sol[1]}}}$

Løsningsformelen for eksponentiallikning på grunnform ser slik ut:

$\\begin{align}e^{x}&=a\\\\ &\\Downarrow \\\\x&=\\ln(a)\\end{align}$

Vi snur på likningen og regner ut:

$\\begin{align}
e^{\\simplify{{c1}x}}-\\var{e1}&=0\\\\
e^{\\simplify{{c1}x}}&=\\var{e1}\\\\
\\var{c1}x &= \\ln(\\var{e1})\\\\
x &=\\dfrac{\\ln(\\var{e1})}{\\var{c1}}\\\\
\\end{align}$

$x=${precround(ln(e1)/c1,2)}

Likningen $\\simplify{{a2}e^(2*x)+{c2}e^(x)+{b2}=0}$ er en andregradslikning med $e^{x}$ som ukjent. Vi bruker da først abc-formelen for å finne $e^{x}$:

$a=\\var{a2},\\;b=\\var{c2},\\;c=\\var{b2}$

$e^{x}=\\dfrac{\\simplify{-{c2}}\\pm\\sqrt{\\simplify[!collectNumbers,timesDot]{{c2}^{2}-4*{a2}*{b2}}}}{2\\cdot\\var{a2}}=\\dfrac{\\simplify{-{c2}}\\pm\\sqrt{\\simplify{{c2}^2-4*{a2}*{b2}}}}{\\simplify{2*{a2}}}$.

Vi to løsninger for $e^{x}$: $e^{x}= \\var{x2[0]}\\text{ og } e^{x}=\\var{x2[1]}$.

$e^{x}= \\var{x2[0]}\\Rightarrow x=\\ln(\\var{x2[0]})\\approx\\underline{\\underline{\\var{sol2[0]}}}$

$e^{x}= \\var{x2[1]}\\Rightarrow x=\\ln(\\var{x2[1]})\\approx\\underline{\\underline{\\var{sol2[1]}}}$

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$\\simplify{{a1} ln(x)+{b1}={c1}}$

$x=$[[0]]

$\\simplify{{a2}{logn}^{2}+{b2}{logn}+{c2}=0}$

(sett inn laveste verdi først:)

$x=$[[0]] $\\vee$ $x=$[[1]]

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$e^{\\var{c1}x}-\\var{e1}=0$

$x=$ [[0]]

$\\simplify{{a2}e^(2*x)+{c2}e^(x)+{b2}=0}$

$x=$ [[0]] $\\vee$ $x=$ [[1]]

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