Isolating x in Rate of Reaction?

[mwall]

I am working on a problem relating to rate of reaction. I am not sure how to isolate the x in the following equation.

e^-45/(8.31)(x)
e^-45/(8.31)(353)

mwall

 

[mwall]

Sorry, I forgot the rest of the equation.

.072 = e^-(45/8.31*x)
.002 e^-(45/8.31*352)

mwall

 

[uart]

Whenever you're not sure how to proceed with rearranging an expression like that one the first thing you should do is to "fold" the constants together. If you do so then it is simply expressed as,

$$e^{ax} = b$$.

So obviously you just need to take logs of both sides to get,

$$x = \frac{\log(b)}{a}$$

 

[da_willem]

the first thing you should do is to "fold" the constants together

This can be done by remembering $$e^a / e^b = e^{a-b}$$

 

[mwall]

Thanks for your help.

mwall