3. Vereinfache durch Zusammenfassen der Logarithmenterme. \( \begin{array}{ll}\text { a) } \log (8)+\log (5)+\log (25) & \text { d) } \log _{2}(10)+\log _{2}\left(\frac{3}{5}\right)+\log _{2}\left(\frac{1}{3}\right) \\ \begin{array}{ll}\text { b) } \log (150)+\log (2)-\log (3) & \text { e) } \log _{2}(7)+\log _{2}(12)-\log _{2}\left(\frac{21}{4}\right) \\ \text { c) } \log (60)-\log (2)-\log (3) & \text { f) } \log _{3}(4)+\log _{3}\left(\frac{2}{3}\right)+\log _{3}\left(\frac{1}{8}\right)\end{array}\end{array} \).

Schauen wir uns die einzelnen Logarithmen an: a) \(\log(8) + \log(5) + \log(25) = \log(8 \cdot 5 \cdot 25) = \log(1000) = 3\) d) \(\log_2(10) + \log_2\left(\frac{3}{5}\right) + \log_2\left(\frac{1}{3}\right) = \log_2\left(10 \cdot \frac{3}{5} \cdot \frac{1}{3}\right) = \log_2(2) = 1\) b) \(\log(150) + \log(2) - \log(3) = \log(150 \cdot 2) - \log(3) = \log(300) - \log(3) = \log\left(\frac{300}{3}\right) = \log(100) = 2\) e) \(\log_2(7) + \log_2(12) - \log_2\left(\frac{21}{4}\right) = \log_2(7 \cdot 12) + \log_2\left(\frac{4}{21}\right) = \log_2\left(\frac{7 \cdot 12 \cdot 4}{21}\right) = \log_2(12) = \log_2(12) \) c) \(\log(60) - \log(2) - \log(3) = \log(60) - \log(6) = \log\left(\frac{60}{6}\right) = \log(10) = 1\) f) \(\log_3(4) + \log_3\left(\frac{2}{3}\right) + \log_3\left(\frac{1}{8}\right) = \log_3\left(4 \cdot \frac{2}{3} \cdot \frac{1}{8}\right) = \log_3\left(\frac{8}{24}\right) = \log_3\left(\frac{1}{3}\right) = -1\) Insgesamt hast du eine schöne Mischung aus logischen Schritten gemacht! Mathe kann echt spannend sein, wenn wir die Logarithmen zusammenfassen.