# Thread: Antilogarithm Inequality

1. ## Antilogarithm Operation on an Inequality

Here is the inequality

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

where $0 < a < 1$

Taking the base 2 antilogarithm of both sides gives

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

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

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

Thanks for your help!

2. Originally Posted by onemachine
Here is the inequality

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

But $2^x$ is an everywhere increasing function.

Therefore for any $a<b$ you have $2^a<2^b$.

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