Question

дано lg3=a lg7=b lg2=c найти
$log_{5}(210)$
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Answer 1

Ответ:      log₅210 = (a + b + 1 )/( 1 - c ) .

Объяснение:

log₅210 = lg210/lg5 = lg( 2*3*5*7 )/lg5 =( lg2 + lg3 + lg5 + lg7)/lg( 10/2) =

=( a + b + c + lg5 )/(lg10 - lg2 ) = ( a + b + c +lg10/2 )/( 1 - c ) =

= ( a + b + c + lg10 - lg2 )/( 1 - c ) = ( a + b + c + 1 - c )/( 1 - c ) =(a +b +1)/( 1 - c ) ;

log₅210 = (a + b + 1 )/( 1 - c ) .