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psi9epsilon
 2 years ago
Best ResponseYou've already chosen the best response.0is it log 2x or is it log x to the base 2, and also log 23 or log 3 to base 2

psi9epsilon
 2 years ago
Best ResponseYou've already chosen the best response.0please retype the question

nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.1\[\huge \log_2 x  \log_2 3 = 2\]this i guess

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

nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.1\[\huge \log_2 x = 2+ \log_2 3\]\[\huge 2^{\log_2 x} = 2^{2 + \log_2 3}\]\[\huge x = 2^2 \times 2 ^{\log_2 3}\]\[x = 4 \times 3\]

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

nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.1o.O which one, how to type it or how to do it?

nphuongsun93
 2 years ago
Best 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 \[\huge 2^{2+\log_2 3} = 2^2 \times 2 ^{\log_2 3} = 4 \times 3\]
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