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Question Number 32885 by scientist last updated on 05/Apr/18

Answered by MJS last updated on 05/Apr/18

$$\log u=a$$$$\log v=b$$$$\log s=c$$$$\log t=d$$$$\mathrm{all}\mathrm{we}\mathrm{need}\mathrm{is}\log \frac{p}{q}=\log p-\log q$$$$(a-b)(c-d)=(a-c)(b-d)+(a-d)(c-b)$$$$(a-b)(c-d)=ab-bc-ad+cd+ac-ab-cd+bd$$$$(a-b)(c-d)=ac-bc-ad+bd$$$$\mathrm{true}$$

Commented by scientist last updated on 05/Apr/18

$$isnotcleartome.pleasedotheworkingsclearly$$

Commented by MJS last updated on 05/Apr/18

$${\mathrm{what}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{clear}?$$$$1.\mathrm{I}\mathrm{used}a,b,c,d\mathrm{instead}\mathrm{of}$$$$\log u,\log v,\log s,\log t\mathrm{to}\mathrm{save}$$$$\mathrm{time}\mathrm{and}\mathrm{space}$$$$2.\log \frac{p}{q}=\log p-\log q\mathrm{is}\mathrm{true}$$$$\mathrm{for}\mathrm{all}p,q>0\Rightarrow \log \frac{u}{v}=\log u-\log v...$$$$\mathrm{the}\mathrm{rest}\mathrm{is}\mathrm{just}\mathrm{multiplicating}...$$

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