Question

If a^3 + b^3 = 0 and a + b != 0, prove that log (a + b) = 1/2(log a+ log b+log 3)
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Answer 1

$\star\:\:\:\bf\large\underline\blue{Question:-}$

• If $a^{3}+b^{3}=0$ and $a+b\neq0$, prove that $log(a+b)=\frac{1}{2}(log\:a + log\:b + log\:3)$

$\star\:\:\:\bf\large\underline\blue{Proof:-}$

Given that,

${a}^{3} + {b}^{3} = 0 \: ....(i)$

And,

$a + b \neq0$

After factorisation of equation (i), we get,

$\implies (a + b)( {a}^{2} - ab + {b}^{2} ) = 0$

By zero product rule, we can say that,

Either $(a+b)=0$ or $a^{2}-ab+b^{2}=0$.

But, it's given that,

$a + b \neq 0$

So,

${a}^{2} - ab + {b}^{2} = 0$

$\implies {a}^{2} + {b}^{2} = ab$

$\implies {a}^{2} + {b}^{2} + 2ab = ab + 2ab$

$\implies {(a + b)}^{2} = 3ab$

Now, taking log on the both side, we get,

$log(a + b)^{2} = log(3ab)$

Using the property, i.e.,

$log(xy) = log(x) + log(y)$

We get,

$log(a + b)^{2} = log \: a + log \: b + log \: 3$

Now, using the property, i.e.,

$log(x)^{y} = y \: log(x)$

We get,

$2 log(a + b) = ( log \: a + log \: b + log \: 3)$

Dividing both side by 2, we get,

$log(a + b) = \frac{1}{2} ( log \: a + log \: b + log \: 3)$

So,

$log(a + b) = \frac{1}{2} ( log \: a + log \: b + log \: 3)$

$\bf\underline\blue{Hence\:Proved.}$