Question

If a=log12 base 24,b=log24 base 36,c=log 36 base 48,Then show that 1+abc=2bc

Answer 1

SOLUTION

GIVEN

$\sf{a = log_{24}(12) \: , \: b = log_{36}(24) \:, \: c = log_{48}(36) }$

TO PROVE

$\sf{1 + abc = 2bc}$

PROOF

LHS

$\sf{ = 1 + abc }$

$\sf{ = 1 + log_{24}(12) \times log_{36}(24) \times log_{48}(36) }$

$\displaystyle \sf{ = 1 + \frac{ log(12) }{ log(24) } \times \frac{ log(24) }{ log(36) } \times \frac{ log(36) }{ log(48) }}$

$\displaystyle \sf{ = 1 + \frac{ log(12) }{ log(48) } }$

$\displaystyle \sf{ = 1 + log_{48}(12) }$

$\displaystyle \sf{ = log_{48}(48) + log_{48}(12) }$

$\displaystyle \sf{ = log_{48}(48 \times 12) }$

$\displaystyle \sf{ = log_{48}(4 \times 12 \times 12) }$

$\displaystyle \sf{ = log_{48}( {2}^{2} \times {12}^{2} ) }$

$\displaystyle \sf{ = log_{48}( {24}^{2} ) }$

$\displaystyle \sf{ = 2log_{48}( {24} ) }$

RHS

$\displaystyle \sf{ = 2log_{36}( {24} ) \times log_{48}( 36) }$

$\displaystyle \sf{ = 2 \times \frac{ log(24) }{ log(36) } \times \frac{ log(36) }{ log(48) } }$

$\displaystyle \sf{ = 2 \times \frac{ log(24) }{ log(48) } }$

$\displaystyle \sf{ = 2 log_{48}(24) }$

Thus LHS = RHS

Hence proved

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