# Thread: How am I supposed to know that ln9 - ln1 = 2*ln3 ?

1. ## How am I supposed to know that ln9 - ln1 = 2*ln3 ?

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!

2. Originally Posted by s3a
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!
From the basics of logs you're expected to know that $\log_b(1) = 0 \: \: \: , \: \: b > 1$. This is because $b^0 = 1$

This expression is simplified using the log power law: $\ln(a^k) = k\, ln(a) \: \: , \: \: a > 0$ and the knowledge of the fact that $9=3^2$

Thus $\ln(9) - \ln(1) = \ln(3^2) - 0 = 2\ln(3)$

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Laws of logarithms. $b > 1$ in all cases and $a,c > 0$ in all cases

$\log_b(a) + \log_b(c) = \log_b(ac)$

$\log_b(a) - log_b(c) = \log_b \left(\frac{a}{c}\right)$

$\log_b(b) = 1$

$\log_b(a^k) = k\,\log_b(a)$

$\log_b \left(\frac{1}{a^k}\right) = -k\,\log_b(a)$

$\log_b(1) = 0$

$\log_b(a) = c \: \: \rightarrow \: \: a = b^c$

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# from the product of [4/3l2mn] (-5ln2) and 1/3l2mn subtract the product of [7/3l3mn2]and [2/3l2n2] solution

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