Question Number 40873 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]](Q40873.png)
$${If}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{0},\:\:{prove}\:{that}\:\mathrm{log}\:\left({a}+{b}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{a}\:+\mathrm{log}\:{b}\:+\mathrm{log}\:\mathrm{3}\right) \\ $$$$\left[{given}\:{a}+{b}\neq\mathrm{0}\right] \\ $$
Answered by math1967 last updated on 29/Jul/18
![a^3 +b^3 =0 (a+b)(a^2 −ab+b^2 )=0 a^2 −ab+b^2 =0 [∵ (a+b)≠0] a^2 +b^2 =ab (a+b)^2 =3ab 2log(a+b)=log3ab ∴log(a+b)=(1/2)(loga+logb+log3)](Q40899.png)
$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{0} \\ $$$$\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} =\mathrm{0}\:\:\:\:\:\:\:\:\left[\because\:\left({a}+{b}\right)\neq\mathrm{0}\right] \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={ab} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{3}{ab} \\ $$$$\mathrm{2}{log}\left({a}+{b}\right)={log}\mathrm{3}{ab} \\ $$$$\therefore{log}\left({a}+{b}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({loga}+{logb}+{log}\mathrm{3}\right) \\ $$
Commented by FaHmYaSmA last updated on 27/Oct/18
how it can be 3ab ? where '3' come from ?
Commented by math1967 last updated on 27/Oct/18
![a^2 +b^2 =ab a^2 +b^2 +2ab=ab+2ab [add 2ab both side] ∴(a+b)^2 =3ab Is it clear?](Q46496.png)
$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={ab} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}={ab}+\mathrm{2}{ab}\:\left[{add}\:\mathrm{2}{ab}\:{both}\:{side}\right] \\ $$$$\:\therefore\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{3}{ab}\: \\ $$$${Is}\:{it}\:{clear}? \\ $$