elisegayezm

2022-09-29

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

blikath3

Beginner2022-09-30Added 7 answers

Firstly, fixing the typo in the question: $\mathrm{log}(2)\cong 0.3$.

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

$=-2\mathrm{log}(2)\cong -2\times 0.3=-0.6$

Incidentally, do you know why $\mathrm{log}(2)\cong 0.3$?

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

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

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

$=-2\mathrm{log}(2)\cong -2\times 0.3=-0.6$

Incidentally, do you know why $\mathrm{log}(2)\cong 0.3$?

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

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

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