Question

Prove that :-

$\tt \displaystyle{ \sum \limits^{n}_{i = 1} \bigg( \log(a_{i}) \bigg) = \log \bigg( \prod \limits^{n} _{i = 1 } (a_{i})\bigg)}$




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Answer 1

$\large\underline{\sf{Given \:Question - }}$

Prove that,

$\rm :\longmapsto\:\tt \displaystyle{ \sum \limits^{n}_{i = 1} \bigg( \log(a_{i}) \bigg) = \log \bigg( \prod \limits^{n} _{i = 1 } (a_{i})\bigg)}$

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\:\tt \displaystyle{ \sum \limits^{n}_{i = 1} \bigg( \log(a_{i}) \bigg)}$

can be rewritten as

$\rm \: = \: \: log \: a_{1}+ log \: a_{2} + log \: a_{3} + - - - + log \: a_{n}$

We know that,

$\boxed{ \bf{ logx \: + \: logy \: = \: logxy \: }}$

[ See the attachment for the proof of this identity ]

So, using this identity, we can rewrite the above as

$\rm \: = \: \: log \: \bigg(a_{1} \times a_{2} \times a_{3} \times - - - - \times a_{n}\bigg)$

$\rm \: = \: \:{ \log \bigg( \prod \limits^{n} _{i = 1 } (a_{i})\bigg)}$

Hence,

$\boxed{ \qquad \rm{ \tt \displaystyle{ \sum \limits^{n}_{i = 1} \bigg( \log(a_{i}) \bigg) = \log \bigg( \prod \limits^{n} _{i = 1 } (a_{i})\bigg) \qquad}}}$

Additional Information :

$\boxed{ \rm{ {e}^{logx} = x}}$

$\boxed{ \rm{ {e}^{ylogx} = {x}^{y} }}$

$\boxed{ \rm{ {a}^{ log_{a}(x) } = x}}$

$\boxed{ \rm{ {a}^{y log_{a}(x) } = {x}^{y} }}$

$\boxed{ \rm{ log_{x}(x) = 1}}$

$\boxed{ \rm{ log_{ {x}^{y} }( {x}^{z} ) = \frac{z}{y} }}$

$\boxed{ \rm{ log_{ {x}^{y} }( {w}^{z} ) = \frac{z}{y} \: log_{x}(w) }}$

$\boxed{ \rm{ {x}^{y} = z \: \implies \: y \: = \: log_{x}(z) }}$

$\boxed{ \rm{ log \: {e}^{x} \: = \: x \: }}$