Posted by sue on Tuesday, May 31, 2011 at 9:15pm.
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?

math  sue, Tuesday, May 31, 2011 at 9:19pm
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)

math  MathMate, Tuesday, May 31, 2011 at 10:40pm
To find an unknown (x), you need an equation.
What has been supplied is an expression containing x. Does it equal some value?

math  sue, Tuesday, May 31, 2011 at 11:00pm
its actually a physics question
q=(2 coul)[1.56x10^20]^([0.50/sec][t])
t=time in sec.

math  sue, Tuesday, May 31, 2011 at 11:10pm
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

math  MathMate, Tuesday, May 31, 2011 at 11:18pm
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.
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