Exponential system of eqn's.

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
Luke53
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Exponential system of eqn's.

Postby Luke53 » 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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stapel_eliz
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Postby stapel_eliz » Tue Jun 04, 2013 6:51 pm

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...
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:

Luke53
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Re: Exponential system of eqn's.

Postby Luke53 » 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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stapel_eliz
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Postby stapel_eliz » Thu Jun 06, 2013 6:52 pm

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:

Luke53
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Re: Exponential system of eqn's.

Postby Luke53 » 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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