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psi9epsilonBest ResponseYou've already chosen the best response.0
is it log 2x or is it log x to the base 2, and also log 23 or log 3 to base 2
 one year ago

psi9epsilonBest ResponseYou've already chosen the best response.0
please retype the question
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.1
\[\huge \log_2 x  \log_2 3 = 2\]this i guess
 one year ago

UMULASBest ResponseYou've already chosen the best response.0
I mean as usual logs i didd x  = 2 3 But the answer is not 6.
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.1
\[\huge \log_2 x = 2+ \log_2 3\]\[\huge 2^{\log_2 x} = 2^{2 + \log_2 3}\]\[\huge x = 2^2 \times 2 ^{\log_2 3}\]\[x = 4 \times 3\]
 one year ago

UMULASBest ResponseYou've already chosen the best response.0
\_/ what how did you get he log and the 2 together?
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.1
o.O which one, how to type it or how to do it?
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.1
\[\huge \log_2 x = 2 + \log_2 3\]then 2^(everything)\[\huge 2^{\log_2 x} = 2^{2+\log_2 3}\]\[\huge 2^{\log_2 x} = x\]and the other side using a^(m+n) = a^m * a^n \[\huge 2^{2+\log_2 3} = 2^2 \times 2 ^{\log_2 3} = 4 \times 3\]
 one year ago
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