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April 29th, 2014, 04:11 PM  #1 
Newbie Joined: Apr 2014 From: AQ Posts: 3 Thanks: 0 Math Focus: casual math user  Exponential function inequality
Hi! This is something that came up in a practical problem. 9  2^2x >= 0 This was my attempt at solving it with my rusty skillz: 9  2^2x >= 0 2^2x >= 9 2^2x =< 9 (multiplying by 1, swapping) 2^x =< sqrt(9) (hope this is valid) 2^x =< 3 So far so good (I think...?). Now for the last step, isolating x by turning it into a log function. Considering: a^y = x equivalent to loga(x) = y we get something like 2^x =< 3 equivalent to log2(3) =< x, but that doesn't seem to make sense. I should swap the inequality again... (log2(3) >= x, or ~1,585 >= x, which seems to be the answer I want) Why do I have to swap the inequality again? Also, did I do anything wrong in the process? 
April 29th, 2014, 04:33 PM  #2 
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176 
$\displaystyle 2^x\leq3 \Leftrightarrow \ \log_2 2^x\leq\log_2 3\Leftrightarrow x\leq \log_2 3.$

April 29th, 2014, 04:58 PM  #3 
Newbie Joined: Apr 2014 From: AQ Posts: 3 Thanks: 0 Math Focus: casual math user 
Right, that actually makes sense with the breaking down. Thanks!


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exponential, function, inequality 
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