Antilogarithm Inequality

onemachine

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Feb 2, 2012
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Antilogarithm Operation on an Inequality

Here is the inequality

\(\displaystyle \left( 1 - a \right) < 1\),

where \(\displaystyle 0 < a < 1\)

Taking the base 2 antilogarithm of both sides gives

\(\displaystyle 2^{\left( 1 - a \right)} < 2\)

I am not sure if this is antilogarithm operation is allowed because the quanitity \(\displaystyle \left( 1 - a \right)\) is less than one.

P.S. I am not trying to solve this for \(\displaystyle a\). I am just using this as an example so I can figure this out.

Thanks for your help!
 
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Here is the inequality

\(\displaystyle \left( 1 - a \right) < 1\),
where \(\displaystyle 0 < a < 1\)
Taking the base 2 antilogarithm of both sides gives
\(\displaystyle 2^{\left( 1 - a \right)} < 2\)
I am not sure if this is antilogarithm operation is allowed because the quanitity \(\displaystyle \left( 1 - a \right)\) is less than one.
P.S. I am not trying to solve this for \(\displaystyle a\). I am just using this as an example so I can figure this out.
I not sure what antilogarithm means.

But \(\displaystyle 2^x\) is an everywhere increasing function.

Therefore for any \(\displaystyle a<b\) you have \(\displaystyle 2^a<2^b\).
 
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