Question

prove that 1/log2 8+1/log4 8+1/log64 8=3​

Answer 1

Step-by-step explanation:

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Answer 2

SOLUTION

TO PROVE

$\displaystyle \sf \frac{1}{ log_{2}(8) } + \frac{1}{ log_{4}(8) } + \frac{1}{ log_{64}(8) } = 3$

FORMULA TO BE IMPLEMENTED

We are aware of the formula on logarithm that

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

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

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

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

EVALUATION

LHS

$\displaystyle \sf = \frac{1}{ log_{2}(8) } + \frac{1}{ log_{4}(8) } + \frac{1}{ log_{64}(8) }$

$\displaystyle \sf = log_{8}(2) + log_{8}(4) + log_{8}(64)$

$\displaystyle \sf = log_{8}(2 \times 4) + log_{8}( {8}^{2} )$

$\displaystyle \sf = log_{8}(8) + 2 log_{8}(8 )$

$\displaystyle \sf = 1 + (2 \times 1)$

$\displaystyle \sf = 1 + 2$

$\displaystyle \sf = 3$

= RHS

Hence proved

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