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
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*10^20/6.24150965(16)×10^18 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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