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Question Number 4816 by Kelvin last updated on 15/Mar/16

Commented by prakash jain last updated on 15/Mar/16

log .006−log .0012=log ((.006)/(.0012))=log ((60)/(12))=log 5  log .007−log .00243=log ((.007)/(.00243))=log ((700)/(243))  log .008+log 4000=log (.008×4000)=log 32=5log 2  log .0128=7log 2−log 10000=7log 2−4

$$\mathrm{log}\:.\mathrm{006}−\mathrm{log}\:.\mathrm{0012}=\mathrm{log}\:\frac{.\mathrm{006}}{.\mathrm{0012}}=\mathrm{log}\:\frac{\mathrm{60}}{\mathrm{12}}=\mathrm{log}\:\mathrm{5} \\ $$$$\mathrm{log}\:.\mathrm{007}−\mathrm{log}\:.\mathrm{00243}=\mathrm{log}\:\frac{.\mathrm{007}}{.\mathrm{00243}}=\mathrm{log}\:\frac{\mathrm{700}}{\mathrm{243}} \\ $$$$\mathrm{log}\:.\mathrm{008}+\mathrm{log}\:\mathrm{4000}=\mathrm{log}\:\left(.\mathrm{008}×\mathrm{4000}\right)=\mathrm{log}\:\mathrm{32}=\mathrm{5log}\:\mathrm{2} \\ $$$$\mathrm{log}\:.\mathrm{0128}=\mathrm{7log}\:\mathrm{2}−\mathrm{log}\:\mathrm{10000}=\mathrm{7log}\:\mathrm{2}−\mathrm{4} \\ $$

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