Where is this function larger then zero?

[Ad VanderVen]

Homework Statement

It is given that 0 <A, 0 <a, 0 <b, 0 <c. Under what conditions is then
1-exp (-a * A / (b / c + 1)) greater than zero?

Relevant Equations

1-exp (-a * A / (b / c + 1))

I do not know how to proceed.

 

[PeroK]

Homework Statement:: It is given that 0 <A, 0 <a, 0 <b, 0 <c. Under what conditions is then
1-exp (-a * A / (b / c + 1)) greater than zero?
Relevant Equations:: 1-exp (-a * A / (b / c + 1))

I do not know how to proceed.

You have to make an effort.

 

[Ad VanderVen]

$$1-exp(-a*A/(b/c+1))>0$$
$$exp(-a*A/(b/c+1))<1$$
$$ln(-a*A/(b/c+1))<0$$
$$0<-a*A/(b/c+1) $$ and $$-a*A/(b/c+1)<1$$

 

[PeroK]

$$exp(-a*A/(b/c+1))<1$$
$$ln(-a*A/(b/c+1))<0$$

This doesn't look right.

 

[Ad VanderVen]

Sorry, You are right. It should be
$$-a*A/(b/c+1)<0$$
and therefore
$$a*A/(b/c+1)>0$$
and because 0 < a, 0 < b, 0 < c, 0 < A it follows that
$$a*A/(b/c+1)>0$$
is always true.

Am I correct?

 

[PeroK]

Sorry, You are right. It should be
$$-a*A/(b/c+1)<0$$
and therefore
$$a*A/(b/c+1)>0$$
and because 0<a, 0<b, 0<c, )< A it follows that
$$a*A/(b/c+1)>0$$
is always true.

Am I correct?

Yes.

 

[Ad VanderVen]

@PeroK As usual, thanks a lot.

 

[PeroK]

@PeroK As usual, thanks a lot.

There as a quick way. First, the exponential is of the form $e^{-k}$, where $k > 0$. This is always less than $1$, because $e^k > 1$ for $k > 0$, and $e^{-k} = 1/e^{k}$.