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dpaInc
 2 years ago
Best ResponseYou've already chosen the best response.12^(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

j.giudice13
 2 years ago
Best ResponseYou've already chosen the best response.11/2 isnt an option....

dpaInc
 2 years ago
Best ResponseYou've already chosen the best response.1you 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?

j.giudice13
 2 years ago
Best ResponseYou've already chosen the best response.1umm, yea let me see what that looks like

j.giudice13
 2 years ago
Best ResponseYou've already chosen the best response.1i dont have a calculator :(

j.giudice13
 2 years ago
Best ResponseYou've already chosen the best response.1i have so many windows up lol

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1just one more won't hurt http://www.wolframalpha.com

j.giudice13
 2 years ago
Best ResponseYou've already chosen the best response.1i just need an answer to see if i did this right

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1what did you get? I gave you a calculator...

j.giudice13
 2 years ago
Best ResponseYou've already chosen the best response.1my computer will crash. its old i got .185

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1http://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.

j.giudice13
 2 years ago
Best ResponseYou've already chosen the best response.1is that the right answer

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.10.292... is the "right" answer. If you want the most accurate answer you take the thing with the log.

j.giudice13
 2 years ago
Best ResponseYou've already chosen the best response.1thank you very much. simple is all i need

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1:) you are welcome
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