i have an inequality

e^{a}=> e^{b}

can i take take the log of this inequality?

and is the following inequality correct forevery "a" and "b"?

log (e^{a}) => log (e^{b}) then a => b

thanks a lot.

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- Feb 26th 2013, 12:10 AMhastiinequality and logarithm
i have an inequality

e^{a}=> e^{b}

can i take take the log of this inequality?

and is the following inequality correct for**every "a" and "b"**?

log (e^{a}) => log (e^{b}) then a => b

thanks a lot. - Feb 26th 2013, 03:31 AMHallsofIvyRe: inequality and logarithm
Yes, both logarithm and exponential are increasing functions so if $\displaystyle e^a\ge e^b$ then $\displaystyle a\ge b$.

(It would be better to use ">=" for "greater than or equal to". "=>" looks too much like "implies".)