Question

logb^a.logc^b.logd^c=logd^a​

Answer 1

SOLUTION

TO PROVE

$\displaystyle \sf{ log_{b}(a). log_{c}(b).log_{d}(c) = log_{d}(a)}$

FORMULA TO BE IMPLEMENTED

We are aware of the formula on logarithm that

$\sf{1. \: \: \: log( {a}^{n} ) = n log(a) }$

$\sf{2. \: \: log(ab) = log(a) + log(b) }$

$\displaystyle \sf{3. \: \: log \bigg( \frac{a}{b} \bigg) = log(a) - log(b) }$

$\sf{4. \: \: log_{a}(a) = 1}$

$\displaystyle \sf{ 5. \: \: log_{b}(a) = \frac{ log(a) }{ log(b) }}$

EVALUATION

We have to prove that

$\displaystyle \sf{ log_{b}(a). log_{c}(b).log_{d}(c) = log_{d}(a)}$

LHS

$\displaystyle \sf{ = log_{b}(a). log_{c}(b).log_{d}(c) }$

$\displaystyle \sf{ = \frac{ log(a) }{ log(b) } . \frac{ log(b) }{ log(c) }. log_{d}(c) } \: \: \: \: (\: \: by \: formula \: \: 5)$

$\displaystyle \sf{ = \frac{ log(a) }{ log(c) }. log_{d}(c) } \:$

$\displaystyle \sf{ = \frac{ log(a) }{ log(c) } . \frac{ log(c) }{ log(d) } } \: \: \: \: (\: \: by \: formula \: \: 5)$

$\displaystyle \sf{ = \frac{ log(a) }{ log(d) } }$

$\displaystyle \sf{ = log_{d}(a) }$

= RHS

Hence proved

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Answer 2

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