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dpaIncBest ResponseYou've already chosen the best response.1
2^(1 + 2x )= 3 take the log_2 of both sides: 1+2x = log_2 3 subtract 1 from both sides: 2x = (log_2 3)  1 divide by 2 x = [(log_2 3)  1]/2
 one year ago

j.giudice13Best ResponseYou've already chosen the best response.1
1/2 isnt an option....
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
you didn't specify the way you wanted the solution... and 1/2 is not what I came up with. it's x = [(log_2 3)  1]/2 did you want a calculator answer?
 one year ago

j.giudice13Best ResponseYou've already chosen the best response.1
umm, yea let me see what that looks like
 one year ago

j.giudice13Best ResponseYou've already chosen the best response.1
i dont have a calculator :(
 one year ago

j.giudice13Best ResponseYou've already chosen the best response.1
i have so many windows up lol
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
just one more won't hurt http://www.wolframalpha.com
 one year ago

j.giudice13Best ResponseYou've already chosen the best response.1
i just need an answer to see if i did this right
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
what did you get? I gave you a calculator...
 one year ago

j.giudice13Best ResponseYou've already chosen the best response.1
my computer will crash. its old i got .185
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=%5B%28log_2+3%29++1%5D%2F2 if that doesn't work use a calculator from the applications section. Those have been there since time immemorial.
 one year ago

j.giudice13Best ResponseYou've already chosen the best response.1
is that the right answer
 one year ago

TomLikesPhysicsBest ResponseYou've already chosen the best response.1
0.292... is the "right" answer. If you want the most accurate answer you take the thing with the log.
 one year ago

j.giudice13Best ResponseYou've already chosen the best response.1
thank you very much. simple is all i need
 one year ago

TomLikesPhysicsBest ResponseYou've already chosen the best response.1
:) you are welcome
 one year ago
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