Question Number 40872 by scientist last updated on 28/Jul/18
![If a^3 +b^3 =0, prove that log (a+b)=(1/2)(log a +log b +log 3) [given a+b≠0]](https://www.tinkutara.com/question/Q40872.png)
$${If}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{0},\:\:{prove}\:{that}\:\mathrm{log}\:\left({a}+{b}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{a}\:+\mathrm{log}\:{b}\:+\mathrm{log}\:\mathrm{3}\right) \\ $$$$\left[{given}\:{a}+{b}\neq\mathrm{0}\right] \\ $$
Commented by MrW3 last updated on 29/Jul/18
$${I}\:{don}'{t}\:{think}\:{there}\:{are}\:{such}\:{real}\:{values} \\ $$$${for}\:{a}\:{and}\:{b}. \\ $$$$ \\ $$$${such}\:{that}\:\mathrm{log}\:\left({a}+{b}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{a}\:+\mathrm{log}\:{b}\:+\mathrm{log}\:\mathrm{3}\right) \\ $$$${is}\:{defined},\:{we}\:{have}\:{a}>\mathrm{0},\:{b}>\mathrm{0},\:{then} \\ $$$${we}\:{get}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} >\mathrm{0},\:{but}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{0}. \\ $$
Commented by maxmathsup by imad last updated on 29/Jul/18
$${you}\:{are}\:{right}\:{sir}\:{its}\:{a}\:{complex}\:{logarithme}\:{because}\:{a}^{\mathrm{2}} −{ab}\:+{b}^{\mathrm{2}} >\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jul/18
$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${given}\:{a}+{b}\:{not}\:{equals}\:{to}\:\mathrm{0} \\ $$$${so}\:{a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}−{ab}=\mathrm{0} \\ $$$${a}+{b}=\sqrt{\mathrm{3}{ab}}\: \\ $$$${ln}\left({a}+{b}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({lna}+{lnb}+{ln}\mathrm{3}\right) \\ $$$$ \\ $$