How am I supposed to know that ln9 - ln1 = 2*ln3 ? For example, I would know that 9/4 - 3/2 is 15/4 because I can convert 3/2 into 6/4 but how do I such a thing with logarithms?
Any input would be greatly appreciated!
Thanks in advance!
How am I supposed to know that ln9 - ln1 = 2*ln3 ? For example, I would know that 9/4 - 3/2 is 15/4 because I can convert 3/2 into 6/4 but how do I such a thing with logarithms?
Any input would be greatly appreciated!
Thanks in advance!
From the basics of logs you're expected to know that $\displaystyle \log_b(1) = 0 \: \: \: , \: \: b > 1$. This is because $\displaystyle b^0 = 1$
This expression is simplified using the log power law: $\displaystyle \ln(a^k) = k\, ln(a) \: \: , \: \: a > 0$ and the knowledge of the fact that $\displaystyle 9=3^2$
Thus $\displaystyle \ln(9) - \ln(1) = \ln(3^2) - 0 = 2\ln(3)$
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Laws of logarithms. $\displaystyle b > 1$ in all cases and $\displaystyle a,c > 0$ in all cases
$\displaystyle \log_b(a) + \log_b(c) = \log_b(ac)$
$\displaystyle \log_b(a) - log_b(c) = \log_b \left(\frac{a}{c}\right)$
$\displaystyle \log_b(b) = 1$
$\displaystyle \log_b(a^k) = k\,\log_b(a)$
$\displaystyle \log_b \left(\frac{1}{a^k}\right) = -k\,\log_b(a)$
$\displaystyle \log_b(1) = 0$
$\displaystyle \log_b(a) = c \: \: \rightarrow \: \: a = b^c$