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 ${\mathrm{log}}_{5}(2)$?

Robaffonadorkdh
2022-10-02
Answered

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 ${\mathrm{log}}_{5}(2)$?

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Caiden Brewer

Answered 2022-10-03
Author has **5** answers

I woul change base of the logarithm using the property:

${\mathrm{log}}_{b}x=\frac{{\mathrm{log}}_{a}x}{{\mathrm{log}}_{a}(b)}$

in your case:

${\mathrm{log}}_{5}(2)=\frac{\mathrm{log}2}{\mathrm{log}(5)}=\frac{0.03}{0.7}=\frac{3}{100}\ast \frac{10}{7}=\frac{3}{70}=0.043$

Although I am not completely sure about your estimates (in particular 0.03)....the result should give you 0.43 so that ${5}^{0.43}=2!$

${\mathrm{log}}_{b}x=\frac{{\mathrm{log}}_{a}x}{{\mathrm{log}}_{a}(b)}$

in your case:

${\mathrm{log}}_{5}(2)=\frac{\mathrm{log}2}{\mathrm{log}(5)}=\frac{0.03}{0.7}=\frac{3}{100}\ast \frac{10}{7}=\frac{3}{70}=0.043$

Although I am not completely sure about your estimates (in particular 0.03)....the result should give you 0.43 so that ${5}^{0.43}=2!$

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