Question

prove log 32 base 2 =5 using formula log mn base a = log m base a +log n base a​

Answer 1

I hope it will help you

Answer 2

SOLUTION

TO PROVE

$\sf{ log_{2}(32) = 5 \: }$

FORMULA TO BE IMPLEMENTED

$\sf{1. \: \: \: log_{a}(mn) = log_{a}(m) + log_{a}(n) \: }$

$\sf{2. \: \: \: log_{a}(a) = 1 \: }$

PROOF

$\sf{ log_{2}(32) }$

$= \sf{ log_{2}(2 \times 16) }$

$= \sf{ log_{2}(2 ) + log_{2}(16) } \: (by \: formula \: 1)$

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

$= \sf{ log_{2}(2 ) + log_{2}(2) + log_{2}(16) } \: (by \: formula \: 1)$

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

$= \sf{2 log_{2}(2 ) + log_{2}(2 \times 4) } \:$

$= \sf{2 log_{2}(2 ) + log_{2}(2) + log_{2}( 4) } \: (by \: formula \: 1)$

$= \sf{3 log_{2}(2 ) + log_{2}( 4) } \:$

$= \sf{3 log_{2}(2 ) + log_{2}( 2 \times 2) } \:$

$= \sf{3 log_{2}(2 ) + log_{2}(2) + log_{2}( 2) } \: (by \: formula \: 1)$

$= \sf{5 log_{2}(2 ) } \:$

$= \sf{(5 \times 1) } \: \: \: (by \: formula \: 2)$

$= \sf{5}$

Therefore

$\sf{ log_{2}(32) = 5 \: }$

Hence proved

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