X=Y e(to the power of) p x t

Y and p are constant

determine approx values of y and p

when

X = 75 t = 2.25

X = 50 t = 1.25

cheers for your help

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- March 8th 2006, 02:40 PMmathsisfunAlgebra
X=Y e(to the power of) p x t

Y and p are constant

determine approx values of y and p

when

X = 75 t = 2.25

X = 50 t = 1.25

cheers for your help - March 8th 2006, 03:20 PMThePerfectHackerQuote:

Originally Posted by**mathsisfun**

- March 8th 2006, 10:41 PMRich B.
Hi:

From the given information, we can substitute real values for x and t, giving the equations 75 = ye^2.25p and 50 = ye^1.25p. Dividing the first equation by the second eliminates the constant y, and we have 75/e^2.25p = 50/e^1.25p. Cross multiply and solve for p. Finally, substitute the value of p, just determined, into either of the two equations derived in the step #1, and solve for y. I leave the algebra to you. Hint: The product of p and y is between 12 and 13.

Regards,

Rich B. - March 8th 2006, 11:20 PMticbolQuote:

Originally Posted by**mathsisfun**

75 = y[e^(p*2.25)

75/y = e^(2.25p)

Take the natural logs of both sides,

ln(75/y) = (2.25p)ln(e)

ln(75) -ln(y) = 2.25p -----------------(1)

When x=50 and t=1.25,

50 = y[e^(p*1.25)

50/y = e^(1.25p)

Take the natural logs of both sides,

ln(50) -ln(y) = 1.25p -----------------(2)

Let's solve for p first,

(1) minus (2),

ln(75) -ln(50) = 2.25p -1.25p

ln(75/50) = 1.00p

p = ln(3/2) = ln(1.5) ----------------------***

Substitute that into, say, (1),

ln(75) -ln(y) = 2.25ln(1.5)

ln(75) -ln(1.5^(2.25)) = ln(y)

ln[75/(1.5^(2.25))] = ln(y)

y = 75/(1.5^(2.25)) = 30.12 -------***

Check those against (2),

ln(50) -ln(30.12) =? 1.25(ln(1.5))

3.912 -3.405 =? 0.507

0.507 =? 0.507

Yes, so, OK.

Therefore, y=30.12 and p=ln(1.5). -----------answer.

---------------------

Reminders:

----log(a/b) = log(a) -log(b).

----if log(a) = log(b), then a=b.

----c*log(a) = log(a^c).

----if ln(a) = b, then a = e^b, that is why ln(e) = 1, or log(10) = 1. - March 9th 2006, 09:29 AMmathsisfun
Thankyou very much for your help.

I understand now, thanks for the reminders too, this is where I was going wrong so they are very useful.

Cheers