onemachine
New member
- Joined
- Feb 2, 2012
- Messages
- 28
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!
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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