C chutch82 New member Joined Aug 12, 2011 Messages 10 Aug 26, 2011 #1 find the value of x when: x= 5 log base2 (1/16)+8 log base3 (27) I'm a little confused how to do this because I can't put it in my calculator because it's not log base 10.
find the value of x when: x= 5 log base2 (1/16)+8 log base3 (27) I'm a little confused how to do this because I can't put it in my calculator because it's not log base 10.
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,325 Aug 26, 2011 #2 You'll need these: \(\displaystyle 2^{4} = 16\) \(\displaystyle 3^{3} = 27\)
burakaltr Junior Member Joined Aug 26, 2011 Messages 126 Aug 27, 2011 #3 chutch82 said: find the value of x when: x= 5 log base2 (1/16)+8 log base3 (27) I'm a little confused how to do this because I can't put it in my calculator because it's not log base 10. Click to expand... x = 5* log base2 ( 2**-4) + 8* log base3 ( 3**3) x= 5* (-4) + 8* (3) x=-20 + 24 x= 4
chutch82 said: find the value of x when: x= 5 log base2 (1/16)+8 log base3 (27) I'm a little confused how to do this because I can't put it in my calculator because it's not log base 10. Click to expand... x = 5* log base2 ( 2**-4) + 8* log base3 ( 3**3) x= 5* (-4) + 8* (3) x=-20 + 24 x= 4
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,325 Aug 27, 2011 #4 Maybe one day we'll find out if the OP can struggle through a problem. Not today, I guess.
