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September 17, 2014

Posted by **Hannah** on Monday, April 23, 2012 at 12:17pm.

I need to solve for x. I was going to do .50 X 3.20e-1 and then take the square root but this is not correct. I am not sure what to do. Thank you for your help.

Math(Please help) - Reiny, Tuesday, April 17, 2012 at 3:18pm

You probably meant

(.50 - x)/x = 3.2 e^-1= 3.2/e

crossmultiply ...

.50 - x = (3.2/e)x

(3.2/e)x + x = .5

x(3.2/e + 1)

x = .5/(3.2/e + 1)

multiply by e/e

= .5e/(3.2 + e)

multiply by 10/10

= 5e/(32 + 10e)

So I have to multiply 5e by 10 which would be 50e and then multiply (32+10e) by 10?

Math(Please respond) - bobpursley, Wednesday, April 18, 2012 at 7:17pm

Hannah: Professor Reiny solved it for you.

x= 10e/(32+10e)

calculate that. Or if you cant, put this in your google search window

10*e/(32+10*e) =

I understand. If the number was 2.49e-1 I would do the same thing, so it would be 10*e / (249 + 10*e) = ?. Is this correct?

- Chemistry -
**DrBob222**, Monday, April 23, 2012 at 1:11pmI started to write this a couple of days ago when I saw this post but didn't. Frankly, I thought both professor Reiny and professor Pursley misinterpreted the question. I still think so. I don't think it has anything to do with e. I believe you meant

[(0.50-x)/x] = 3.20E-1 which I would rewrite as

0.320 = (0.50-x)/x, then

0.320x = 0.50-x

1.320x = 0.50 and

x = 0.50/1.320

x = ?

- Chemistry -
**Hannah**, Monday, April 23, 2012 at 1:33pmOk thank you the way you have it is exactly the way that it was stated on my homework. The only problem is that I realized that I had the typed the wrong number. It was suppose to be the same equation but with 2.49e-1 instead 3.20e-1. I understand everything except how did you get 1.320?

- Chemistry -
**DrBob222**, Monday, April 23, 2012 at 2:22pm0.320 = (0.50-x)/x, then

0.320x = 0.50-x**0.320x = 0.50 -x**

add x to both sides.

0.320x + x = 0.50 -x +x

0.320x + 1x = 0.50

1.320x = 0.50 and

x = 0.50/1.320.

1.320x = 0.50 and

x = 0.50/1.320

- Chemistry -
**Hannah**, Monday, April 23, 2012 at 2:27pmThank you.

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