Question Number 32885 by scientist last updated on 05/Apr/18
Answered by MJS last updated on 05/Apr/18
$$\mathrm{log}\:{u}={a} \\ $$$$\mathrm{log}\:{v}={b} \\ $$$$\mathrm{log}\:{s}={c} \\ $$$$\mathrm{log}\:{t}={d} \\ $$$$\mathrm{all}\:\mathrm{we}\:\mathrm{need}\:\mathrm{is}\:\mathrm{log}\:\frac{{p}}{{q}}=\mathrm{log}\:{p}−\mathrm{log}\:{q} \\ $$$$\left({a}−{b}\right)\left({c}−{d}\right)=\left({a}−{c}\right)\left({b}−{d}\right)+\left({a}−{d}\right)\left({c}−{b}\right) \\ $$$$\left({a}−{b}\right)\left({c}−{d}\right)={ab}−{bc}−{ad}+{cd}+{ac}−{ab}−{cd}+{bd} \\ $$$$\left({a}−{b}\right)\left({c}−{d}\right)={ac}−{bc}−{ad}+{bd} \\ $$$$\mathrm{true} \\ $$
Commented by scientist last updated on 05/Apr/18
$${is}\:{not}\:{clear}\:{to}\:{me}.\:{please}\:{do}\:{the}\:{workings}\:{clearly} \\ $$
Commented by MJS last updated on 05/Apr/18
$$\mathrm{what}'\mathrm{s}\:\mathrm{not}\:\mathrm{clear}? \\ $$$$\mathrm{1}.\:\mathrm{I}\:\mathrm{used}\:{a},\:{b},\:{c},\:{d}\:\mathrm{instead}\:\mathrm{of} \\ $$$$\mathrm{log}\:{u},\:\mathrm{log}\:{v},\:\mathrm{log}\:{s},\:\mathrm{log}\:{t}\:\mathrm{to}\:\mathrm{save} \\ $$$$\mathrm{time}\:\mathrm{and}\:\mathrm{space} \\ $$$$\mathrm{2}.\:\mathrm{log}\:\frac{{p}}{{q}}=\mathrm{log}\:{p}−\mathrm{log}\:{q}\:\mathrm{is}\:\mathrm{true} \\ $$$$\mathrm{for}\:\mathrm{all}\:{p},\:{q}>\mathrm{0}\:\Rightarrow\:\mathrm{log}\:\frac{{u}}{{v}}=\mathrm{log}\:{u}−\mathrm{log}\:{v}... \\ $$$$\mathrm{the}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{just}\:\mathrm{multiplicating}... \\ $$