2024-10-30 21:21:19 +01:00
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#import "@preview/grape-suite:1.0.0": exercise
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#import exercise: project, task, subtask
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#set text(lang: "de")
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#show: project.with(
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title: "Mathe am 28.10.2024",
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seminar: [Mathe Q2],
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show-outline: true,
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author: "Erik Grobecker",
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// date: 28.10.2024,
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show-solutions: false
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)
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#show math.equation: set text(font: "New Computer Modern Math")
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= Hausaufgaben
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1. Potenzregelen wiederholen & lernen
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2. S. 109 Nr. 1 d) bis j) machen
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== Potenzregelen
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== S. 109 Nr. 1 d) bis j)
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*d)*\
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$
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3 dot e^(4x) &= 16.2\
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16.2/3 &approx 5.4\
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ln(root(4, 5.4)) &approx 0.42\
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ln(root(4, 16.2/3))&approx 0.42
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$
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2024-11-04 12:32:17 +01:00
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Schrittweise:
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2024-10-30 21:21:19 +01:00
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$
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3 dot e^(4x) &= 16.2 &&|:3\
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e^(4x)&=5.4 &&|ln()\
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4x&approx 1.69 &&|:4\
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x& approx 0.421
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$
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*e)*\
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$
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e^(-x)&=10\
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ln(e^(-x))&=ln(10)\
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-x dot -1 &approx 2.3 dot -1\
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x &= -2.3
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$
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*f)*\
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$
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e^(4-x)&=1\
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ln(e^(4-x))&=ln(1)\
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4-x&=0
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x&=4
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$
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2024-11-04 12:32:17 +01:00
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#pagebreak()
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2024-10-30 21:21:19 +01:00
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*g)*\
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$
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e^(4-4x)&=5\
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ln(e^(4-4x))&=ln(5)\
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(4-4x)/4 &approx 1.61/4\
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1-x-1 &approx 0.4 -1\
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-x dot (-1) &=-0.6 dot -1\
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x &= 0.6
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$
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*h)*\
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$
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2e^(-x)&=5\
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(2e^(-x))/2 &=5/2\
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-ln((2e^(-x))/2)&=-ln(5/2) approx -0.92
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$
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*i)*\
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$
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e^(2x+1)&=10\
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ln(e^(2x+1))&=ln(10)\
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(ln(e^(2x+1))-1)/2 &=(ln(10)-1)/2 approx 0.65
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$
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*j)*\
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$
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3 dot e^(0.5x-1)&=1\
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(3 dot e^(0.5x-1))/3 &= 1/3\
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ln((3 dot e^(0.5x-1))/3) &= ln(1/3)\
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ln((3 dot e^(0.5x-1))/3)+1 &= ln(1/3)+1\
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(ln((3 dot e^(0.5x-1))/3)+1) dot 2 &= (ln(1/3)+1) dot 2 approx -0.197
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$
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