lehrkraefte:ks:miniex:ex05 [https://fginfo.ksbg.ch Fachgruppe Informatik der KSBG]

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<JS>
function generate(jQuery, idex, idsol, ex, sep="<br>", sep2="<br>", numex=3) {
    var randperm=function(n) {
	var a = [];
	for (var i=0; i<n; i++) { a[i]=i; }
	for (var i=0; i<n; i++) {
	    var j = Math.floor(Math.random()*(n-i))+i;
	    if (j>i) {
		var h = a[j];
		a[j] = a[i];
		a[i] = h;
	    }
	}
	return a
    }
    var selec=randperm(ex.length);
    if (numex<1){
	numex = ex.length;
    }
    for (var i=0; i<numex; i++) {
       jQuery(idex).append((i+1)+". &nbsp; "+ex[selec[i]][0]+sep);
       jQuery(idsol).append((i+1)+". &nbsp; "+ex[selec[i]][1]+sep2);
    }
}
</JS>
==== 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:
<JS>jQuery(function() {generate(jQuery, "#exoexpgleichung","#solexpgleichung",
[["$\\left(\\frac{1}{256}\\right)^{\\frac{2}{5}x+\\frac{1}{2}}=\\frac{1}{1024}$", "$\\begin{align*}\n\\left(\\frac{1}{256}\\right)^{\\frac{2}{5}x+\\frac{1}{2}}&=\\frac{1}{1024}&&|\\log_{2}(\\cdot)\\\\\n\\log_{2}\\left(\\left(\\frac{1}{256}\\right)^{\\frac{2}{5}x+\\frac{1}{2}}\\right)&=\\log_{2}\\left(2^{-10}\\right)\\\\\n\\left(\\frac{2}{5}x+\\frac{1}{2}\\right)\\cdot \\log_{2}\\left(2^{-8}\\right) & = -10\\\\\n\\left(\\frac{2}{5}x+\\frac{1}{2}\\right)\\cdot -8 & = -10 && |:-8\\\\\n\\frac{2}{5}x+\\frac{1}{2} & = \\frac{5}{4} && | -\\frac{1}{2}\\\\\n\\frac{2}{5}x & = \\frac{3}{4} && |:\\frac{2}{5}\\\\\nx & = \\frac{15}{8}\n\\end{align*}$"], ["$64^{-\\frac{4}{3}x+\\frac{5}{2}}=\\frac{1}{128}$", "$\\begin{align*}\n64^{-\\frac{4}{3}x+\\frac{5}{2}}&=\\frac{1}{128}&&|\\log_{2}(\\cdot)\\\\\n\\log_{2}\\left(64^{-\\frac{4}{3}x+\\frac{5}{2}}\\right)&=\\log_{2}\\left(2^{-7}\\right)\\\\\n\\left(-\\frac{4}{3}x+\\frac{5}{2}\\right)\\cdot \\log_{2}\\left(2^{6}\\right) & = -7\\\\\n\\left(-\\frac{4}{3}x+\\frac{5}{2}\\right)\\cdot 6 & = -7 && |:6\\\\\n-\\frac{4}{3}x+\\frac{5}{2} & = -\\frac{7}{6} && | -\\frac{5}{2}\\\\\n-\\frac{4}{3}x & = -\\frac{11}{3} && |:-\\frac{4}{3}\\\\\nx & = \\frac{11}{4}\n\\end{align*}$"], ["$81^{-\\frac{5}{6}x+\\frac{4}{3}}=\\frac{1}{27}$", "$\\begin{align*}\n81^{-\\frac{5}{6}x+\\frac{4}{3}}&=\\frac{1}{27}&&|\\log_{3}(\\cdot)\\\\\n\\log_{3}\\left(81^{-\\frac{5}{6}x+\\frac{4}{3}}\\right)&=\\log_{3}\\left(3^{-3}\\right)\\\\\n\\left(-\\frac{5}{6}x+\\frac{4}{3}\\right)\\cdot \\log_{3}\\left(3^{4}\\right) & = -3\\\\\n\\left(-\\frac{5}{6}x+\\frac{4}{3}\\right)\\cdot 4 & = -3 && |:4\\\\\n-\\frac{5}{6}x+\\frac{4}{3} & = -\\frac{3}{4} && | -\\frac{4}{3}\\\\\n-\\frac{5}{6}x & = -\\frac{25}{12} && |:-\\frac{5}{6}\\\\\nx & = \\frac{5}{2}\n\\end{align*}$"], ["$64^{\\frac{2}{3}x-\\frac{1}{2}}=1024$", "$\\begin{align*}\n64^{\\frac{2}{3}x-\\frac{1}{2}}&=1024&&|\\log_{2}(\\cdot)\\\\\n\\log_{2}\\left(64^{\\frac{2}{3}x-\\frac{1}{2}}\\right)&=\\log_{2}\\left(2^{10}\\right)\\\\\n\\left(\\frac{2}{3}x-\\frac{1}{2}\\right)\\cdot \\log_{2}\\left(2^{6}\\right) & = 10\\\\\n\\left(\\frac{2}{3}x-\\frac{1}{2}\\right)\\cdot 6 & = 10 && |:6\\\\\n\\frac{2}{3}x-\\frac{1}{2} & = \\frac{5}{3} && | +\\frac{1}{2}\\\\\n\\frac{2}{3}x & = \\frac{13}{6} && |:\\frac{2}{3}\\\\\nx & = \\frac{13}{4}\n\\end{align*}$"], ["$25^{\\frac{2}{3}x+\\frac{5}{6}}=\\frac{1}{625}$", "$\\begin{align*}\n25^{\\frac{2}{3}x+\\frac{5}{6}}&=\\frac{1}{625}&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(25^{\\frac{2}{3}x+\\frac{5}{6}}\\right)&=\\log_{5}\\left(5^{-4}\\right)\\\\\n\\left(\\frac{2}{3}x+\\frac{5}{6}\\right)\\cdot \\log_{5}\\left(5^{2}\\right) & = -4\\\\\n\\left(\\frac{2}{3}x+\\frac{5}{6}\\right)\\cdot 2 & = -4 && |:2\\\\\n\\frac{2}{3}x+\\frac{5}{6} & = -2/1 && | -\\frac{5}{6}\\\\\n\\frac{2}{3}x & = -\\frac{17}{6} && |:\\frac{2}{3}\\\\\nx & = -\\frac{17}{4}\n\\end{align*}$"], ["$1024^{-\\frac{2}{3}x-\\frac{3}{5}}=64$", "$\\begin{align*}\n1024^{-\\frac{2}{3}x-\\frac{3}{5}}&=64&&|\\log_{2}(\\cdot)\\\\\n\\log_{2}\\left(1024^{-\\frac{2}{3}x-\\frac{3}{5}}\\right)&=\\log_{2}\\left(2^{6}\\right)\\\\\n\\left(-\\frac{2}{3}x-\\frac{3}{5}\\right)\\cdot \\log_{2}\\left(2^{10}\\right) & = 6\\\\\n\\left(-\\frac{2}{3}x-\\frac{3}{5}\\right)\\cdot 10 & = 6 && |:10\\\\\n-\\frac{2}{3}x-\\frac{3}{5} & = \\frac{3}{5} && | +\\frac{3}{5}\\\\\n-\\frac{2}{3}x & = \\frac{6}{5} && |:-\\frac{2}{3}\\\\\nx & = -\\frac{9}{5}\n\\end{align*}$"], ["$\\left(\\frac{1}{625}\\right)^{-\\frac{1}{2}x-\\frac{1}{3}}=\\frac{1}{125}$", "$\\begin{align*}\n\\left(\\frac{1}{625}\\right)^{-\\frac{1}{2}x-\\frac{1}{3}}&=\\frac{1}{125}&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(\\left(\\frac{1}{625}\\right)^{-\\frac{1}{2}x-\\frac{1}{3}}\\right)&=\\log_{5}\\left(5^{-3}\\right)\\\\\n\\left(-\\frac{1}{2}x-\\frac{1}{3}\\right)\\cdot \\log_{5}\\left(5^{-4}\\right) & = -3\\\\\n\\left(-\\frac{1}{2}x-\\frac{1}{3}\\right)\\cdot -4 & = -3 && |:-4\\\\\n-\\frac{1}{2}x-\\frac{1}{3} & = \\frac{3}{4} && | +\\frac{1}{3}\\\\\n-\\frac{1}{2}x & = \\frac{13}{12} && |:-\\frac{1}{2}\\\\\nx & = -\\frac{13}{6}\n\\end{align*}$"], ["$27^{-\\frac{5}{2}x+\\frac{5}{3}}=\\frac{1}{81}$", "$\\begin{align*}\n27^{-\\frac{5}{2}x+\\frac{5}{3}}&=\\frac{1}{81}&&|\\log_{3}(\\cdot)\\\\\n\\log_{3}\\left(27^{-\\frac{5}{2}x+\\frac{5}{3}}\\right)&=\\log_{3}\\left(3^{-4}\\right)\\\\\n\\left(-\\frac{5}{2}x+\\frac{5}{3}\\right)\\cdot \\log_{3}\\left(3^{3}\\right) & = -4\\\\\n\\left(-\\frac{5}{2}x+\\frac{5}{3}\\right)\\cdot 3 & = -4 && |:3\\\\\n-\\frac{5}{2}x+\\frac{5}{3} & = -\\frac{4}{3} && | -\\frac{5}{3}\\\\\n-\\frac{5}{2}x & = -3/1 && |:-\\frac{5}{2}\\\\\nx & = \\frac{6}{5}\n\\end{align*}$"], ["$625^{-\\frac{2}{5}x-\\frac{1}{2}}=25$", "$\\begin{align*}\n625^{-\\frac{2}{5}x-\\frac{1}{2}}&=25&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(625^{-\\frac{2}{5}x-\\frac{1}{2}}\\right)&=\\log_{5}\\left(5^{2}\\right)\\\\\n\\left(-\\frac{2}{5}x-\\frac{1}{2}\\right)\\cdot \\log_{5}\\left(5^{4}\\right) & = 2\\\\\n\\left(-\\frac{2}{5}x-\\frac{1}{2}\\right)\\cdot 4 & = 2 && |:4\\\\\n-\\frac{2}{5}x-\\frac{1}{2} & = \\frac{1}{2} && | +\\frac{1}{2}\\\\\n-\\frac{2}{5}x & = 1/1 && |:-\\frac{2}{5}\\\\\nx & = -\\frac{5}{2}\n\\end{align*}$"], ["$8^{\\frac{1}{2}x+\\frac{3}{5}}=\\frac{1}{64}$", "$\\begin{align*}\n8^{\\frac{1}{2}x+\\frac{3}{5}}&=\\frac{1}{64}&&|\\log_{2}(\\cdot)\\\\\n\\log_{2}\\left(8^{\\frac{1}{2}x+\\frac{3}{5}}\\right)&=\\log_{2}\\left(2^{-6}\\right)\\\\\n\\left(\\frac{1}{2}x+\\frac{3}{5}\\right)\\cdot \\log_{2}\\left(2^{3}\\right) & = -6\\\\\n\\left(\\frac{1}{2}x+\\frac{3}{5}\\right)\\cdot 3 & = -6 && |:3\\\\\n\\frac{1}{2}x+\\frac{3}{5} & = -2/1 && | -\\frac{3}{5}\\\\\n\\frac{1}{2}x & = -\\frac{13}{5} && |:\\frac{1}{2}\\\\\nx & = -\\frac{26}{5}\n\\end{align*}$"]],
" <hr> ", " <hr> ");});
</JS>
<HTML>
<div id="exoexpgleichung"></div>

</HTML>
<hidden Lösungen>
<HTML>
<div id="solexpgleichung"></div>
</HTML>
</hidden>

<JS>
function generate(jQuery, idex, idsol, ex, sep="<br>", sep2="<br>") {
    var drei=function(n) {
      var one = Math.floor(Math.random()*(n-2));  
      var two = Math.floor(Math.random()*(n-2));
      if (one>two) {
          var h = one;
          one = two;
          two = h;
      }
      var three = Math.floor(Math.random()*(n-2));
      if (two>=one) {
          two++;
      }
      if (three>=one) {
          three++;
      }
      if (three>=two) {
          three++;
      }
      return [one,two,three];
    };
    
    var selec=drei(ex.length);
    //console.log(selec);
    
    for (var i=0; i<3; i++) {
       //console.log(selec[i]);
       jQuery(idex).append((i+1)+". &nbsp; "+ex[selec[i]][0]+sep);
       jQuery(idsol).append((i+1)+". &nbsp; "+ex[selec[i]][1]+sep2);
    }
    
}
</JS>








==== Aufgabe 3 ====
Berechne it Hilfe eines geeigneten Basiswechsels:
<JS>jQuery(function() {generate(jQuery, "#exobasiswechsel","#solbasiswechsel",
[["$\\log_{25}(\\frac{1}{625})$", "$\\log_{25}(\\frac{1}{625})$=$\\frac{\\log_{5}\\left(5^{-4}\\right)}{\\log_{5}\\left(5^{2}\\right)}=\\frac{-4}{2}=-\\frac{2}{1}$"], ["$\\log_{\\frac{1}{32}}(256)$", "$\\log_{\\frac{1}{32}}(256)$=$\\frac{\\log_{2}\\left(2^{8}\\right)}{\\log_{2}\\left(2^{-5}\\right)}=\\frac{8}{-5}=-\\frac{8}{5}$"], ["$\\log_{\\frac{1}{8}}(32)$", "$\\log_{\\frac{1}{8}}(32)$=$\\frac{\\log_{2}\\left(2^{5}\\right)}{\\log_{2}\\left(2^{-3}\\right)}=\\frac{5}{-3}=-\\frac{5}{3}$"], ["$\\log_{\\frac{1}{512}}(16)$", "$\\log_{\\frac{1}{512}}(16)$=$\\frac{\\log_{2}\\left(2^{4}\\right)}{\\log_{2}\\left(2^{-9}\\right)}=\\frac{4}{-9}=-\\frac{4}{9}$"], ["$\\log_{64}(\\frac{1}{128})$", "$\\log_{64}(\\frac{1}{128})$=$\\frac{\\log_{2}\\left(2^{-7}\\right)}{\\log_{2}\\left(2^{6}\\right)}=\\frac{-7}{6}=-\\frac{7}{6}$"], ["$\\log_{\\frac{1}{8}}(512)$", "$\\log_{\\frac{1}{8}}(512)$=$\\frac{\\log_{2}\\left(2^{9}\\right)}{\\log_{2}\\left(2^{-3}\\right)}=\\frac{9}{-3}=-\\frac{3}{1}$"], ["$\\log_{\\frac{1}{8}}(128)$", "$\\log_{\\frac{1}{8}}(128)$=$\\frac{\\log_{2}\\left(2^{7}\\right)}{\\log_{2}\\left(2^{-3}\\right)}=\\frac{7}{-3}=-\\frac{7}{3}$"], ["$\\log_{\\frac{1}{512}}(64)$", "$\\log_{\\frac{1}{512}}(64)$=$\\frac{\\log_{2}\\left(2^{6}\\right)}{\\log_{2}\\left(2^{-9}\\right)}=\\frac{6}{-9}=-\\frac{2}{3}$"], ["$\\log_{64}(\\frac{1}{512})$", "$\\log_{64}(\\frac{1}{512})$=$\\frac{\\log_{2}\\left(2^{-9}\\right)}{\\log_{2}\\left(2^{6}\\right)}=\\frac{-9}{6}=-\\frac{3}{2}$"], ["$\\log_{\\frac{1}{32}}(8)$", "$\\log_{\\frac{1}{32}}(8)$=$\\frac{\\log_{2}\\left(2^{3}\\right)}{\\log_{2}\\left(2^{-5}\\right)}=\\frac{3}{-5}=-\\frac{3}{5}$"]],
" <br> ");});
</JS>
<HTML>
<div id="exobasiswechsel"></div>

</HTML>
<hidden Lösungen>
<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>
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