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--- lehrkraefte:ks:miniex:ex05 [2019/02/16 21:01]
Simon Knaus
+++ lehrkraefte:ks:miniex:ex05 [2019/02/16 21:28] (current)
Simon Knaus [Aufgabe 3]
@@ Line 24: / Line 24: @@
 }
 </JS>
-=== Freitag 2. November 2018 ===
+==== Aufgabe 1 ====
+Berechne die folgenden Audrücke von Hand:
+  - $\log_3(27)$
+  - $\log_{27}(81)$
+  - $\log_{16}(2)$
+  - $\log_{4}(\sqrt{2})$
+<hidden Lösungen>
+Zur Erinnerung: Es ist immer $\log_{a}(b)=c\Leftrightarrow a^c=b$. In Worten: $\log_a(b)$ beantwortet die Frage <<$a$ hoch wie viel gibt $b$?>>.
+  - $3^{3}=27 \Rightarrow \log_3{27}=3$
+  - $27^{\frac{4}{3}}=81\Rightarrow \log_{27}{81}=\frac{4}{3}$
+  - $16^{\frac14}=2\Rightarrow\log_{16}(2)=\frac14$
+  - $4^{\frac{1}{4}}=2^{\frac{1}{2}}\Rightarrow\log_{4}(\sqrt{2})=\frac{1}{4}$
+</hidden>
+==== Aufgabe 2 ====
 Lösen Sie die Gleichung von Hand auf:
@@ Line 83: / Line 98: @@
-==== 22. Januar 2018 bis 26. Januar 2018 ====
+==== Aufgabe 3 ====
-=== Dienstag 23. Januar 2018 und Donnerstag 25. Januar 2018 ===
+Berechne it Hilfe eines geeigneten Basiswechsels:
-**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:
 <JS>jQuery(function() {generate(jQuery, "#exobasiswechsel","#solbasiswechsel",
 [["$\\log_{25}(\\frac{1}{625})$", "$\\log_{25}(\\frac{1}{625})$=$\\frac{\\log_{5}\\left(5^{-4}\\right)}{\\log_{5}\\left(5^{2}\\right)}=\\frac{-4}{2}=-\\frac{2}{1}$"], ["$\\log_{\\frac{1}{32}}(256)$", "$\\log_{\\frac{1}{32}}(256)$=$\\frac{\\log_{2}\\left(2^{8}\\right)}{\\log_{2}\\left(2^{-5}\\right)}=\\frac{8}{-5}=-\\frac{8}{5}$"], ["$\\log_{\\frac{1}{8}}(32)$", "$\\log_{\\frac{1}{8}}(32)$=$\\frac{\\log_{2}\\left(2^{5}\\right)}{\\log_{2}\\left(2^{-3}\\right)}=\\frac{5}{-3}=-\\frac{5}{3}$"], ["$\\log_{\\frac{1}{512}}(16)$", "$\\log_{\\frac{1}{512}}(16)$=$\\frac{\\log_{2}\\left(2^{4}\\right)}{\\log_{2}\\left(2^{-9}\\right)}=\\frac{4}{-9}=-\\frac{4}{9}$"], ["$\\log_{64}(\\frac{1}{128})$", "$\\log_{64}(\\frac{1}{128})$=$\\frac{\\log_{2}\\left(2^{-7}\\right)}{\\log_{2}\\left(2^{6}\\right)}=\\frac{-7}{6}=-\\frac{7}{6}$"], ["$\\log_{\\frac{1}{8}}(512)$", "$\\log_{\\frac{1}{8}}(512)$=$\\frac{\\log_{2}\\left(2^{9}\\right)}{\\log_{2}\\left(2^{-3}\\right)}=\\frac{9}{-3}=-\\frac{3}{1}$"], ["$\\log_{\\frac{1}{8}}(128)$", "$\\log_{\\frac{1}{8}}(128)$=$\\frac{\\log_{2}\\left(2^{7}\\right)}{\\log_{2}\\left(2^{-3}\\right)}=\\frac{7}{-3}=-\\frac{7}{3}$"], ["$\\log_{\\frac{1}{512}}(64)$", "$\\log_{\\frac{1}{512}}(64)$=$\\frac{\\log_{2}\\left(2^{6}\\right)}{\\log_{2}\\left(2^{-9}\\right)}=\\frac{6}{-9}=-\\frac{2}{3}$"], ["$\\log_{64}(\\frac{1}{512})$", "$\\log_{64}(\\frac{1}{512})$=$\\frac{\\log_{2}\\left(2^{-9}\\right)}{\\log_{2}\\left(2^{6}\\right)}=\\frac{-9}{6}=-\\frac{3}{2}$"], ["$\\log_{\\frac{1}{32}}(8)$", "$\\log_{\\frac{1}{32}}(8)$=$\\frac{\\log_{2}\\left(2^{3}\\right)}{\\log_{2}\\left(2^{-5}\\right)}=\\frac{3}{-5}=-\\frac{3}{5}$"]],
@@ Line 99: / Line 111: @@
 <HTML>
 <div id="solbasiswechsel"></div>
+</HTML>
+</hidden>
+==== Auftrag 4 ====
+Berechnen Sie von Hand (Repetieren Sie dazu 2er Potenzen bis $2^{10}$, 3er Potenzen bis $3^4$, damit 4er bis $4^5$ und 5er Potenzen bis $5^4$).
+<JS>jQuery(function() {generate(jQuery, "#exologpot","#sollogpot",
+[["$\\log_{4}\\left(\\frac{1}{1024}\\right)+\\log_{3}\\left(27\\right)+\\log_{4}\\left(\\frac{1}{64}\\right)$", "$-5+3+-3=-5$"], ["$\\log_{5}\\left(\\frac{1}{25}\\right)+\\log_{2}\\left(64\\right)+\\log_{2}\\left(\\frac{1}{1024}\\right)$", "$-2+6+-10=-6$"], ["$\\log_{2}\\left(\\frac{1}{64}\\right)+\\log_{2}\\left(512\\right)+\\log_{5}\\left(\\frac{1}{625}\\right)$", "$-6+9+-4=-1$"], ["$\\log_{4}\\left(256\\right)+\\log_{2}\\left(\\frac{1}{128}\\right)+\\log_{4}\\left(64\\right)$", "$4+-7+3=0$"], ["$\\log_{2}\\left(\\frac{1}{512}\\right)+\\log_{3}\\left(81\\right)+\\log_{2}\\left(\\frac{1}{256}\\right)$", "$-9+4+-8=-13$"], ["$\\log_{3}\\left(\\frac{1}{27}\\right)+\\log_{3}\\left(\\frac{1}{81}\\right)+\\log_{2}\\left(32\\right)$", "$-3+-4+5=-2$"], ["$\\log_{5}\\left(25\\right)+\\log_{5}\\left(625\\right)+\\log_{2}\\left(128\\right)$", "$2+4+7=13$"], ["$\\log_{2}\\left(\\frac{1}{32}\\right)+\\log_{4}\\left(\\frac{1}{256}\\right)+\\log_{2}\\left(256\\right)$", "$-5+-4+8=-1$"], ["$\\log_{5}\\left(\\frac{1}{125}\\right)+\\log_{5}\\left(125\\right)+\\log_{4}\\left(1024\\right)$", "$-3+3+5=5$"]],
+" <br> ");});
+</JS>
+<HTML>
+<div id="exologpot"></div>
+</HTML>
+<hidden Lösungen>
+<HTML>
+<div id="sollogpot"></div>
+</HTML>
+</hidden>
+==== Aufgabe 5 ====
+Schreibe als Summe von Vielfachen von $\log_a(x)$ und $\log_a(y)$. Verwende dabei die Logarithmus Gesetze.
+<JS>jQuery(function() {generate(jQuery, "#exologlaws","#solloglaws",
+[["$\\log_a\\left(\\left(x^{4}y^{-3}\\right)^{5}\\right)+\\log_a\\left(\\left(x^{3}y^{4}\\right)^{4}\\right)$", "$\\begin{multline*}\\log_a\\left(\\left(x^{4}y^{-3}\\right)^{5}\\right)+\\log_a\\left(\\left(x^{3}y^{4}\\right)^{4}\\right)=\\log_a\\left(\\left(x^{4}y^{-3}\\right)^{5} \\cdot \\left(x^{3}y^{4}\\right)^{4}\\right) = \\\\\\log_a\\left(x^{20}y^{-15} \\cdot x^{12}y^{16}\\right) =\\log_a\\left(x^{32} \\cdot y^{1} \\right) = \\\\\\log_a\\left(x^{32}\\right) + \\log_a\\left(y^{1} \\right) = 32 \\log_a(x) +1\\log_a(y)\\end{multline*}$"], ["$\\log_a\\left(\\left(x^{3}y^{4}\\right)^{-3}\\right)+\\log_a\\left(\\left(x^{2}y^{-3}\\right)^{-5}\\right)$", "$\\begin{multline*}\\log_a\\left(\\left(x^{3}y^{4}\\right)^{-3}\\right)+\\log_a\\left(\\left(x^{2}y^{-3}\\right)^{-5}\\right)=\\log_a\\left(\\left(x^{3}y^{4}\\right)^{-3} \\cdot \\left(x^{2}y^{-3}\\right)^{-5}\\right) = \\\\\\log_a\\left(x^{-9}y^{-12} \\cdot x^{-10}y^{15}\\right) =\\log_a\\left(x^{-19} \\cdot y^{3} \\right) = \\\\\\log_a\\left(x^{-19}\\right) + \\log_a\\left(y^{3} \\right) = -19 \\log_a(x) +3\\log_a(y)\\end{multline*}$"], ["$\\log_a\\left(\\left(x^{5}y^{-5}\\right)^{5}\\right)+\\log_a\\left(\\left(x^{-2}y^{-4}\\right)^{-5}\\right)$", "$\\begin{multline*}\\log_a\\left(\\left(x^{5}y^{-5}\\right)^{5}\\right)+\\log_a\\left(\\left(x^{-2}y^{-4}\\right)^{-5}\\right)=\\log_a\\left(\\left(x^{5}y^{-5}\\right)^{5} \\cdot \\left(x^{-2}y^{-4}\\right)^{-5}\\right) = \\\\\\log_a\\left(x^{25}y^{-25} \\cdot x^{10}y^{20}\\right) =\\log_a\\left(x^{35} \\cdot y^{-5} \\right) = \\\\\\log_a\\left(x^{35}\\right) + \\log_a\\left(y^{-5} \\right) = 35 \\log_a(x) -5\\log_a(y)\\end{multline*}$"], ["$\\log_a\\left(\\left(x^{3}y^{-3}\\right)^{-5}\\right)+\\log_a\\left(\\left(x^{4}y^{-4}\\right)^{2}\\right)$", "$\\begin{multline*}\\log_a\\left(\\left(x^{3}y^{-3}\\right)^{-5}\\right)+\\log_a\\left(\\left(x^{4}y^{-4}\\right)^{2}\\right)=\\log_a\\left(\\left(x^{3}y^{-3}\\right)^{-5} \\cdot \\left(x^{4}y^{-4}\\right)^{2}\\right) = \\\\\\log_a\\left(x^{-15}y^{15} \\cdot x^{8}y^{-8}\\right) =\\log_a\\left(x^{-7} \\cdot y^{7} \\right) = \\\\\\log_a\\left(x^{-7}\\right) + \\log_a\\left(y^{7} \\right) = -7 \\log_a(x) +7\\log_a(y)\\end{multline*}$"], ["$\\log_a\\left(\\left(x^{5}y^{4}\\right)^{5}\\right)+\\log_a\\left(\\left(x^{3}y^{-4}\\right)^{2}\\right)$", "$\\begin{multline*}\\log_a\\left(\\left(x^{5}y^{4}\\right)^{5}\\right)+\\log_a\\left(\\left(x^{3}y^{-4}\\right)^{2}\\right)=\\log_a\\left(\\left(x^{5}y^{4}\\right)^{5} \\cdot \\left(x^{3}y^{-4}\\right)^{2}\\right) = \\\\\\log_a\\left(x^{25}y^{20} \\cdot x^{6}y^{-8}\\right) =\\log_a\\left(x^{31} \\cdot y^{12} \\right) = \\\\\\log_a\\left(x^{31}\\right) + \\log_a\\left(y^{12} \\right) = 31 \\log_a(x) +12\\log_a(y)\\end{multline*}$"], ["$\\log_a\\left(\\left(x^{3}y^{-3}\\right)^{-5}\\right)+\\log_a\\left(\\left(x^{2}y^{3}\\right)^{4}\\right)$", "$\\begin{multline*}\\log_a\\left(\\left(x^{3}y^{-3}\\right)^{-5}\\right)+\\log_a\\left(\\left(x^{2}y^{3}\\right)^{4}\\right)=\\log_a\\left(\\left(x^{3}y^{-3}\\right)^{-5} \\cdot \\left(x^{2}y^{3}\\right)^{4}\\right) = \\\\\\log_a\\left(x^{-15}y^{15} \\cdot x^{8}y^{12}\\right) =\\log_a\\left(x^{-7} \\cdot y^{27} \\right) = \\\\\\log_a\\left(x^{-7}\\right) + \\log_a\\left(y^{27} \\right) = -7 \\log_a(x) +27\\log_a(y)\\end{multline*}$"], ["$\\log_a\\left(\\left(x^{5}y^{-5}\\right)^{2}\\right)+\\log_a\\left(\\left(x^{4}y^{5}\\right)^{5}\\right)$", "$\\begin{multline*}\\log_a\\left(\\left(x^{5}y^{-5}\\right)^{2}\\right)+\\log_a\\left(\\left(x^{4}y^{5}\\right)^{5}\\right)=\\log_a\\left(\\left(x^{5}y^{-5}\\right)^{2} \\cdot \\left(x^{4}y^{5}\\right)^{5}\\right) = \\\\\\log_a\\left(x^{10}y^{-10} \\cdot x^{20}y^{25}\\right) =\\log_a\\left(x^{30} \\cdot y^{15} \\right) = \\\\\\log_a\\left(x^{30}\\right) + \\log_a\\left(y^{15} \\right) = 30 \\log_a(x) +15\\log_a(y)\\end{multline*}$"], ["$\\log_a\\left(\\left(x^{-3}y^{-2}\\right)^{-2}\\right)+\\log_a\\left(\\left(x^{-3}y^{4}\\right)^{5}\\right)$", "$\\begin{multline*}\\log_a\\left(\\left(x^{-3}y^{-2}\\right)^{-2}\\right)+\\log_a\\left(\\left(x^{-3}y^{4}\\right)^{5}\\right)=\\log_a\\left(\\left(x^{-3}y^{-2}\\right)^{-2} \\cdot \\left(x^{-3}y^{4}\\right)^{5}\\right) = \\\\\\log_a\\left(x^{6}y^{4} \\cdot x^{-15}y^{20}\\right) =\\log_a\\left(x^{-9} \\cdot y^{24} \\right) = \\\\\\log_a\\left(x^{-9}\\right) + \\log_a\\left(y^{24} \\right) = -9 \\log_a(x) +24\\log_a(y)\\end{multline*}$"], ["$\\log_a\\left(\\left(x^{-4}y^{-5}\\right)^{-2}\\right)+\\log_a\\left(\\left(x^{3}y^{4}\\right)^{3}\\right)$", "$\\begin{multline*}\\log_a\\left(\\left(x^{-4}y^{-5}\\right)^{-2}\\right)+\\log_a\\left(\\left(x^{3}y^{4}\\right)^{3}\\right)=\\log_a\\left(\\left(x^{-4}y^{-5}\\right)^{-2} \\cdot \\left(x^{3}y^{4}\\right)^{3}\\right) = \\\\\\log_a\\left(x^{8}y^{10} \\cdot x^{9}y^{12}\\right) =\\log_a\\left(x^{17} \\cdot y^{22} \\right) = \\\\\\log_a\\left(x^{17}\\right) + \\log_a\\left(y^{22} \\right) = 17 \\log_a(x) +22\\log_a(y)\\end{multline*}$"], ["$\\log_a\\left(\\left(x^{2}y^{-3}\\right)^{3}\\right)+\\log_a\\left(\\left(x^{-3}y^{5}\\right)^{-5}\\right)$", "$\\begin{multline*}\\log_a\\left(\\left(x^{2}y^{-3}\\right)^{3}\\right)+\\log_a\\left(\\left(x^{-3}y^{5}\\right)^{-5}\\right)=\\log_a\\left(\\left(x^{2}y^{-3}\\right)^{3} \\cdot \\left(x^{-3}y^{5}\\right)^{-5}\\right) = \\\\\\log_a\\left(x^{6}y^{-9} \\cdot x^{15}y^{-25}\\right) =\\log_a\\left(x^{21} \\cdot y^{-34} \\right) = \\\\\\log_a\\left(x^{21}\\right) + \\log_a\\left(y^{-34} \\right) = 21 \\log_a(x) -34\\log_a(y)\\end{multline*}$"]],
+" <br> ");});
+</JS>
+<HTML>
+<div id="exologlaws"></div>
+</HTML>
+<hidden Lösungen>
+<HTML>
+<div id="solloglaws"></div>
 </HTML>
 </hidden>