1. ## Base e

Here is an easy one for someone to nail.

Need to solve: In x = -79.93

I know x = e^-76.93

but how do i get to the answer 4.93 x 10^-34

I have an "In" button on my calc., the shift function of this is "e^x".

2. Hello,

Originally Posted by chem_student
Here is an easy one for someone to nail.

Need to solve: In x = -79.93

I know x = e^-76.93

but how do i get to the answer 4.93 x 10^-34

I have an "In" button on my calc., the shift function of this is "e^x".

First of all, it's LN, not IN, for "logarithm neperian".

Then you should have in your calculator (can you tell me which one if you don't find it ?) a button with "e^x", or something alike.
Otherwise, try exp(...)

3. Yes I have "e^x" its the shift function of the Ln key.

So i push shift, then Ln, and the number 1 comes up.

4. Originally Posted by chem_student
Yes I have "e^x" its the shift function of the Ln key.

So i push shift, then Ln, and the number 1 comes up.
That's strange...
What calculator do you have ?

Because if 1 comes up, it means that you typed ln(e)

5. Ok so I punched in -76.93 and then the function "e^x".

Got 3.89 x 10^-34

Teachers answer to the question is 4.93 x 10^-34.

6. Originally Posted by chem_student
Ok so I punched in -76.93 and then the function "e^x".

Got 3.89 x 10^-34

Teachers answer to the question is 4.93 x 10^-34.
Because it's 79.93, not 76.93

7. that gives me 1.94 x 10^35

damn this.

8. I was talking about the part without the sign...

You said in your first message that it was $\displaystyle -79.93$

9. I give up.

Cant get 4.93 x 10^-34 no matter what I do.

Thanks anyway.

10. I think the answer your teacher provided you with is false... I've tried it out and it doesn't yield his answer

Are you sure you copied it well ?

(%i1) %e^(-79.93);
(%o1) 1.9357178794122214*10^-35

(%i2) 10^(-79.93);
(%o2) 1.1748975549394918*10^-80

(%i3) 10^(-76.93);
(%o3) 1.1748975549394913*10^-77

(%i4) %e^(-76.93);
(%o4) 3.8879932939808711*10^-34