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April 29th, 2014, 04:11 PM   #1
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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?
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April 29th, 2014, 04:33 PM   #2
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$\displaystyle 2^x\leq3 \Leftrightarrow \ \log_2 2^x\leq\log_2 3\Leftrightarrow x\leq \log_2 3.$
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April 29th, 2014, 04:58 PM   #3
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Right, that actually makes sense with the breaking down. Thanks!
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