# Based on the estimates log(2) = .03 and log(5) = .7, how do you use properties of logarithms to find approximate values for log(0.25)?

May 22, 2015

Firstly, fixing the typo in the question: $\log \left(2\right) \cong 0.3$.

$\log \left(0.25\right) = \log \left(\frac{1}{4}\right) = \log \left(\frac{1}{2} ^ 2\right) = \log \left({2}^{- 2}\right)$

$= - 2 \log \left(2\right) \cong - 2 \times 0.3 = - 0.6$

Incidentally, do you know why $\log \left(2\right) \cong 0.3$?

$10 \times \log \left(2\right) = \log \left({2}^{10}\right) = \log \left(1024\right) \cong \log \left(1000\right) = \log \left({10}^{3}\right) = 3$

So basically it's because ${2}^{10} \cong {10}^{3}$.