miniaufgabe.js
==== 22. Januar 2018 bis 26. Januar 2018 ====
=== Dienstag 23. Januar 2018 und Donnerstag 25. Januar 2018 ===
**Elternsprechtag** am Freitag Nachmittag, 16. Februar. **Bitte Termine reservieren** (13:00 bis 19:00), schriftlich in der Stunde oder per e-mail.
Berechnen Sie mit Hilfe eines geeigneten Basiswechsels:
miniAufgabe("#exobasiswechsel","#solbasiswechsel",
[["$\\log_{25}(\\frac{1}{625})$", "$\\log_{25}(\\frac{1}{625})$=$\\frac{\\log_{5}\\left(5^{-4}\\right)}{\\log_{5}\\left(5^{2}\\right)}=\\frac{-4}{2}=-\\frac{2}{1}$"], ["$\\log_{\\frac{1}{32}}(256)$", "$\\log_{\\frac{1}{32}}(256)$=$\\frac{\\log_{2}\\left(2^{8}\\right)}{\\log_{2}\\left(2^{-5}\\right)}=\\frac{8}{-5}=-\\frac{8}{5}$"], ["$\\log_{\\frac{1}{8}}(32)$", "$\\log_{\\frac{1}{8}}(32)$=$\\frac{\\log_{2}\\left(2^{5}\\right)}{\\log_{2}\\left(2^{-3}\\right)}=\\frac{5}{-3}=-\\frac{5}{3}$"], ["$\\log_{\\frac{1}{512}}(16)$", "$\\log_{\\frac{1}{512}}(16)$=$\\frac{\\log_{2}\\left(2^{4}\\right)}{\\log_{2}\\left(2^{-9}\\right)}=\\frac{4}{-9}=-\\frac{4}{9}$"], ["$\\log_{64}(\\frac{1}{128})$", "$\\log_{64}(\\frac{1}{128})$=$\\frac{\\log_{2}\\left(2^{-7}\\right)}{\\log_{2}\\left(2^{6}\\right)}=\\frac{-7}{6}=-\\frac{7}{6}$"], ["$\\log_{\\frac{1}{8}}(512)$", "$\\log_{\\frac{1}{8}}(512)$=$\\frac{\\log_{2}\\left(2^{9}\\right)}{\\log_{2}\\left(2^{-3}\\right)}=\\frac{9}{-3}=-\\frac{3}{1}$"], ["$\\log_{\\frac{1}{8}}(128)$", "$\\log_{\\frac{1}{8}}(128)$=$\\frac{\\log_{2}\\left(2^{7}\\right)}{\\log_{2}\\left(2^{-3}\\right)}=\\frac{7}{-3}=-\\frac{7}{3}$"], ["$\\log_{\\frac{1}{512}}(64)$", "$\\log_{\\frac{1}{512}}(64)$=$\\frac{\\log_{2}\\left(2^{6}\\right)}{\\log_{2}\\left(2^{-9}\\right)}=\\frac{6}{-9}=-\\frac{2}{3}$"], ["$\\log_{64}(\\frac{1}{512})$", "$\\log_{64}(\\frac{1}{512})$=$\\frac{\\log_{2}\\left(2^{-9}\\right)}{\\log_{2}\\left(2^{6}\\right)}=\\frac{-9}{6}=-\\frac{3}{2}$"], ["$\\log_{\\frac{1}{32}}(8)$", "$\\log_{\\frac{1}{32}}(8)$=$\\frac{\\log_{2}\\left(2^{3}\\right)}{\\log_{2}\\left(2^{-5}\\right)}=\\frac{3}{-5}=-\\frac{3}{5}$"]],
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=== Freitag 26. Januar 2018 ===
Lösen Sie die Gleichung von Hand auf:
miniAufgabe("#exoexpgleichung","#solexpgleichung",
[["$\\left(\\frac{1}{81}\\right)^{\\frac{1}{3}x+\\frac{5}{2}}=27$", "$\\begin{align*}\n\\left(\\frac{1}{81}\\right)^{\\frac{1}{3}x+\\frac{5}{2}}&=27&&|\\log_{3}(\\cdot)\\\\\n\\log_{3}\\left(\\left(\\frac{1}{81}\\right)^{\\frac{1}{3}x+\\frac{5}{2}}\\right)&=\\log_{3}\\left(3^{3}\\right)\\\\\n\\left(\\frac{1}{3}x+\\frac{5}{2}\\right)\\cdot \\log_{3}\\left(3^{-4}\\right) & = 3\\\\\n\\left(\\frac{1}{3}x+\\frac{5}{2}\\right)\\cdot -4 & = 3 && |:-4\\\\\n\\frac{1}{3}x+\\frac{5}{2} & = -\\frac{3}{4} && | -\\frac{5}{2}\\\\\n\\frac{1}{3}x & = -\\frac{13}{4} && |:\\frac{1}{3}\\\\\nx & = -\\frac{39}{4}\n\\end{align*}$"], ["$\\left(\\frac{1}{27}\\right)^{\\frac{1}{2}x-\\frac{1}{2}}=\\frac{1}{81}$", "$\\begin{align*}\n\\left(\\frac{1}{27}\\right)^{\\frac{1}{2}x-\\frac{1}{2}}&=\\frac{1}{81}&&|\\log_{3}(\\cdot)\\\\\n\\log_{3}\\left(\\left(\\frac{1}{27}\\right)^{\\frac{1}{2}x-\\frac{1}{2}}\\right)&=\\log_{3}\\left(3^{-4}\\right)\\\\\n\\left(\\frac{1}{2}x-\\frac{1}{2}\\right)\\cdot \\log_{3}\\left(3^{-3}\\right) & = -4\\\\\n\\left(\\frac{1}{2}x-\\frac{1}{2}\\right)\\cdot -3 & = -4 && |:-3\\\\\n\\frac{1}{2}x-\\frac{1}{2} & = \\frac{4}{3} && | +\\frac{1}{2}\\\\\n\\frac{1}{2}x & = \\frac{11}{6} && |:\\frac{1}{2}\\\\\nx & = \\frac{11}{3}\n\\end{align*}$"], ["$\\left(\\frac{1}{125}\\right)^{\\frac{3}{2}x+\\frac{4}{3}}=\\frac{1}{25}$", "$\\begin{align*}\n\\left(\\frac{1}{125}\\right)^{\\frac{3}{2}x+\\frac{4}{3}}&=\\frac{1}{25}&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(\\left(\\frac{1}{125}\\right)^{\\frac{3}{2}x+\\frac{4}{3}}\\right)&=\\log_{5}\\left(5^{-2}\\right)\\\\\n\\left(\\frac{3}{2}x+\\frac{4}{3}\\right)\\cdot \\log_{5}\\left(5^{-3}\\right) & = -2\\\\\n\\left(\\frac{3}{2}x+\\frac{4}{3}\\right)\\cdot -3 & = -2 && |:-3\\\\\n\\frac{3}{2}x+\\frac{4}{3} & = \\frac{2}{3} && | -\\frac{4}{3}\\\\\n\\frac{3}{2}x & = -\\frac{2}{3} && |:\\frac{3}{2}\\\\\nx & = -\\frac{4}{9}\n\\end{align*}$"], ["$125^{\\frac{2}{5}x+\\frac{4}{3}}=\\frac{1}{625}$", "$\\begin{align*}\n125^{\\frac{2}{5}x+\\frac{4}{3}}&=\\frac{1}{625}&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(125^{\\frac{2}{5}x+\\frac{4}{3}}\\right)&=\\log_{5}\\left(5^{-4}\\right)\\\\\n\\left(\\frac{2}{5}x+\\frac{4}{3}\\right)\\cdot \\log_{5}\\left(5^{3}\\right) & = -4\\\\\n\\left(\\frac{2}{5}x+\\frac{4}{3}\\right)\\cdot 3 & = -4 && |:3\\\\\n\\frac{2}{5}x+\\frac{4}{3} & = -\\frac{4}{3} && | -\\frac{4}{3}\\\\\n\\frac{2}{5}x & = -\\frac{8}{3} && |:\\frac{2}{5}\\\\\nx & = -\\frac{20}{3}\n\\end{align*}$"], ["$\\left(\\frac{1}{625}\\right)^{\\frac{2}{3}x+\\frac{5}{6}}=\\frac{1}{25}$", "$\\begin{align*}\n\\left(\\frac{1}{625}\\right)^{\\frac{2}{3}x+\\frac{5}{6}}&=\\frac{1}{25}&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(\\left(\\frac{1}{625}\\right)^{\\frac{2}{3}x+\\frac{5}{6}}\\right)&=\\log_{5}\\left(5^{-2}\\right)\\\\\n\\left(\\frac{2}{3}x+\\frac{5}{6}\\right)\\cdot \\log_{5}\\left(5^{-4}\\right) & = -2\\\\\n\\left(\\frac{2}{3}x+\\frac{5}{6}\\right)\\cdot -4 & = -2 && |:-4\\\\\n\\frac{2}{3}x+\\frac{5}{6} & = \\frac{1}{2} && | -\\frac{5}{6}\\\\\n\\frac{2}{3}x & = -\\frac{1}{3} && |:\\frac{2}{3}\\\\\nx & = -\\frac{1}{2}\n\\end{align*}$"], ["$\\left(\\frac{1}{125}\\right)^{-\\frac{1}{2}x-\\frac{1}{2}}=25$", "$\\begin{align*}\n\\left(\\frac{1}{125}\\right)^{-\\frac{1}{2}x-\\frac{1}{2}}&=25&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(\\left(\\frac{1}{125}\\right)^{-\\frac{1}{2}x-\\frac{1}{2}}\\right)&=\\log_{5}\\left(5^{2}\\right)\\\\\n\\left(-\\frac{1}{2}x-\\frac{1}{2}\\right)\\cdot \\log_{5}\\left(5^{-3}\\right) & = 2\\\\\n\\left(-\\frac{1}{2}x-\\frac{1}{2}\\right)\\cdot -3 & = 2 && |:-3\\\\\n-\\frac{1}{2}x-\\frac{1}{2} & = -\\frac{2}{3} && | +\\frac{1}{2}\\\\\n-\\frac{1}{2}x & = -\\frac{1}{6} && |:-\\frac{1}{2}\\\\\nx & = \\frac{1}{3}\n\\end{align*}$"], ["$81^{\\frac{5}{4}x-\\frac{3}{2}}=27$", "$\\begin{align*}\n81^{\\frac{5}{4}x-\\frac{3}{2}}&=27&&|\\log_{3}(\\cdot)\\\\\n\\log_{3}\\left(81^{\\frac{5}{4}x-\\frac{3}{2}}\\right)&=\\log_{3}\\left(3^{3}\\right)\\\\\n\\left(\\frac{5}{4}x-\\frac{3}{2}\\right)\\cdot \\log_{3}\\left(3^{4}\\right) & = 3\\\\\n\\left(\\frac{5}{4}x-\\frac{3}{2}\\right)\\cdot 4 & = 3 && |:4\\\\\n\\frac{5}{4}x-\\frac{3}{2} & = \\frac{3}{4} && | +\\frac{3}{2}\\\\\n\\frac{5}{4}x & = \\frac{9}{4} && |:\\frac{5}{4}\\\\\nx & = \\frac{9}{5}\n\\end{align*}$"], ["$125^{-\\frac{3}{4}x-\\frac{1}{2}}=\\frac{1}{25}$", "$\\begin{align*}\n125^{-\\frac{3}{4}x-\\frac{1}{2}}&=\\frac{1}{25}&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(125^{-\\frac{3}{4}x-\\frac{1}{2}}\\right)&=\\log_{5}\\left(5^{-2}\\right)\\\\\n\\left(-\\frac{3}{4}x-\\frac{1}{2}\\right)\\cdot \\log_{5}\\left(5^{3}\\right) & = -2\\\\\n\\left(-\\frac{3}{4}x-\\frac{1}{2}\\right)\\cdot 3 & = -2 && |:3\\\\\n-\\frac{3}{4}x-\\frac{1}{2} & = -\\frac{2}{3} && | +\\frac{1}{2}\\\\\n-\\frac{3}{4}x & = -\\frac{1}{6} && |:-\\frac{3}{4}\\\\\nx & = \\frac{2}{9}\n\\end{align*}$"], ["$27^{-\\frac{3}{4}x+\\frac{5}{6}}=81$", "$\\begin{align*}\n27^{-\\frac{3}{4}x+\\frac{5}{6}}&=81&&|\\log_{3}(\\cdot)\\\\\n\\log_{3}\\left(27^{-\\frac{3}{4}x+\\frac{5}{6}}\\right)&=\\log_{3}\\left(3^{4}\\right)\\\\\n\\left(-\\frac{3}{4}x+\\frac{5}{6}\\right)\\cdot \\log_{3}\\left(3^{3}\\right) & = 4\\\\\n\\left(-\\frac{3}{4}x+\\frac{5}{6}\\right)\\cdot 3 & = 4 && |:3\\\\\n-\\frac{3}{4}x+\\frac{5}{6} & = \\frac{4}{3} && | -\\frac{5}{6}\\\\\n-\\frac{3}{4}x & = \\frac{1}{2} && |:-\\frac{3}{4}\\\\\nx & = -\\frac{2}{3}\n\\end{align*}$"], ["$81^{\\frac{4}{3}x-\\frac{5}{4}}=27$", "$\\begin{align*}\n81^{\\frac{4}{3}x-\\frac{5}{4}}&=27&&|\\log_{3}(\\cdot)\\\\\n\\log_{3}\\left(81^{\\frac{4}{3}x-\\frac{5}{4}}\\right)&=\\log_{3}\\left(3^{3}\\right)\\\\\n\\left(\\frac{4}{3}x-\\frac{5}{4}\\right)\\cdot \\log_{3}\\left(3^{4}\\right) & = 3\\\\\n\\left(\\frac{4}{3}x-\\frac{5}{4}\\right)\\cdot 4 & = 3 && |:4\\\\\n\\frac{4}{3}x-\\frac{5}{4} & = \\frac{3}{4} && | +\\frac{5}{4}\\\\\n\\frac{4}{3}x & = 2/1 && |:\\frac{4}{3}\\\\\nx & = \\frac{3}{2}\n\\end{align*}$"]],
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