Exponential system of eqn's.  TOPIC_SOLVED

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Exponential system of eqn's.

Postby Luke53 on Tue Jun 04, 2013 5:48 pm

How can I solve this system? (e = natural log)

x * e ^ -(0.2/y) = 4 (1)

x * e ^ -(1/y) = 1 (2)

I know that x= 5.65 and y= .577 but how do I get these values?

From the second eqn. I get:

ln x = 1/y, but how do I solve x = 4 * e ^(.2/y) ?

Thanks.
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Postby stapel_eliz on Tue Jun 04, 2013 6:51 pm

Luke53 wrote:How can I solve this system? (e = natural log)

The natural-log function is denoted by "ln(x)", where "x" is the argument of the log function. Are you perhaps referring to the natural exponential, approximately equal to 2.71828? If so, then...

Luke53 wrote:x * e ^ -(0.2/y) = 4 (1)
x * e ^ -(1/y) = 1 (2)

My understanding of the system is as follows:

. . . . .

. . . . .

You can solve each of these equations for "x=", and then set the results equal, giving you one equation in one variable. Take the natural log of each side, and solve for the value of y. Back-solve for x.

I haven't done the complete solution, but I've gone far enough to confirm the approximate value you've listed for y. :wink:
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Re: Exponential system of eqn's.

Postby Luke53 on Thu Jun 06, 2013 1:53 pm

Sorry, but I still can't solve these eqn's so please help.

This is what I have till now:

ln (4 * e^(.2/y))= ln e ^(1/y)) => (1.386 * .2)/y ln e = (1/y) ln e ( with ln e = 1)

So (1.386 * .2)/y = 1/y .

But these are clearly not equal to each other, so I must have done something wrong, but I can't figure out what.


Thanks

Luke.
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  TOPIC_SOLVED

Postby stapel_eliz on Thu Jun 06, 2013 6:52 pm

Luke53 wrote:This is what I have till now:

ln (4 * e^(.2/y))= ln e ^(1/y)) => (1.386 * .2)/y ln e = (1/y) ln e ( with ln e = 1)

Check your log rules: ln(a*b) does not equal ln(a)*ln(b)! :wink:
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Re: Exponential system of eqn's.

Postby Luke53 on Fri Jun 07, 2013 2:14 pm

OK, the log rules helped me out, I found the solutions now, thanks for your help.

Greetings.

Luke.
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