prove that log125base10=3-3log2base10
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log C base A = log C base B / log A base B
you get :
log 8 base 2 + [ log 8 base 2 / log 4 base 2] + [ log 8 base 2 / log 16 base 2] ;
Now you easily solve all the logs by their definition and get:
= 3 + [ 3 / 2 ] + [ 3 / 4] = (12 + 6 +3)/4 = 21 /4 !
you get :
log 8 base 2 + [ log 8 base 2 / log 4 base 2] + [ log 8 base 2 / log 16 base 2] ;
Now you easily solve all the logs by their definition and get:
= 3 + [ 3 / 2 ] + [ 3 / 4] = (12 + 6 +3)/4 = 21 /4 !
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