Potenz und Logarithmusgesetze

$\displaystyle a^m \cdot a^n = a^{m+n}$

$\displaystyle \dfrac{a^m}{a^n} = a^{m-n}$

$\displaystyle (a^m)^n = a^{m\cdot n}$

$\displaystyle a^0 = 1$

$\displaystyle a^1 = a$

$\displaystyle a^m \cdot b^m = (a\cdot b)^m$

$\displaystyle \dfrac{a^m}{b^m} = \left(\dfrac{a}{b}\right)^m$

$\displaystyle \log_a(b) = x \Leftrightarrow a^x = b$

$\displaystyle \log_a(b) = \dfrac{\log_c(b)}{\log_c(a)}$

$\displaystyle \log_c(a\cdot b) = \log_c(b) + \log_c(a)$

$\displaystyle \log_c\left(\dfrac{a}{b}\right) = \log_c(b) - \log_c(a)$

$\displaystyle \log_c(b^r) = r\cdot \log_c(b)$

Daraus folgen beispiel:

denkanregung:

$\displaystyle \log_{20}{4000} \Leftrightarrow (10\cdot 2)$

Also: auf gut deutsch: versuchen: 2er Logarithmus und 10er Logarithmus zu vereinigen. z.B.

$\displaystyle (2\cdot 10)^x = 4000, 2^x \cdot 10^x, log_20(1000) = log_c(1000)\log_c(20)$

oder:

$\displaystyle 8 = 4\cdot 2 \Leftrightarrow ld(4\cdot 2) = ld(4) + ld(2) = 2+1 = 3, ld(8) = 3$

$\displaystyle 16 = 8\cdot 2 \Leftrightarrow ld(8\cdot 2) = ld(8) + ld(2) = 3+1 = 4, ld(16) = 4$

$\displaystyle 16 = 4\cdot 4 \Leftrightarrow ld(4\cdot 4) = ld(4) + ld(4) = 2+2 = 4, ld(16) = 4$

$\displaystyle 512 = \dfrac{2048}{4} \Leftrightarrow ld\left(\dfrac{2048}{4}\right) = ld(2048) - ld(4) = 11-2 = 9, ld(512) = 9$