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
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.
When x=75 and t=2.25,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.