math
posted by sue .
How do i find X in e^(0.50/x) on a scientific calc?
do I use LN, and if so, could you pls tell me how I can plug it in?

actually the e is a variable with the value 1.56x10^20
So, I have to find X in [1.56x10^20]^(0.50/X) 
To find an unknown (x), you need an equation.
What has been supplied is an expression containing x. Does it equal some value? 
its actually a physics question
q=(2 coul)[1.56x10^20]^([0.50/sec][t])
t=time in sec. 
The full question:
The charge (in coul.) on an object increases according to q=(2 coul)e^[(o.05/sec)(t)], where t=time in sec. At how many sec. is the object deficient 1.56x10^20 electrons (from neutral)?
A 0 B 12.56 C 21.94 D 28 E 42.21 F 50.5 
One coulomb = 6.24150965(16)×10^18 electrons.
1.56*10^20 electrons
= 1.56*10^20/6.24150965(16)×10^18 coulombs
=25 coulombs
So the equation becomes:
25 coulombs = 2 coulombs * e^(0.05t)
e^(0.05t) = 25/2=12.5
take ln on both sides:
0.05t = ln 12.5 = 2.526
t = 2.526/0.05 = 50.5 sec.