1. ## Algebra

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

2. Originally Posted by mathsisfun
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

There is a dictum in math that if the number of variables exceed the number of equations then you cannot find unique solutions. Over here you have two variable you need to solve for p and y. And only one equation.

3. 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.

4. Originally Posted by mathsisfun
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

When x=75 and t=2.25,
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.

---------------------
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.

5. 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