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November 24, 2014

November 24, 2014

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

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 9:19pmactually 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 10:40pmTo 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 11:00pmits 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 11:10pmThe 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 11:18pmOne 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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