How can I solve this?
\(40Xe^{-\frac{5}{2}X}=0.1\)
e, in this context, is Euler's number.
40Xe^((-5/2)X)=0.1
How can I solve this?
e, in this context, is Euler's number.
40Xe^((-5/2)X)=0.1
X = 0.00251577
X = 2.80984
This type of a problem can be solved by either "graphing" it, or a sure way, which is what I did, is "Interpolation".
Assuming that you wanted to do this with just the aid of a pocket calculator, you would use a numerical method.
Probably best, would be the Newton-Raphson method.
Check the 'Great Answers to Learn From' in Sticky Topics, there's bound to be an example there.
An alternative would be to use a simple fixed point iteration.
Rewrite the equation as
exp(-5x/2) = 0.1/40x = 0.0025/x.
take logs
-5x/2 = ln(0.0025/x),
so
x = (-2/5)ln(0.0025/x) = -0.4ln(0.0025/x).
Now choose a starting value for x, substitute that into the rhs and that will give you a new value for x.
Substitute that into the rhs and repeat the routine (as many times as you need in order to get the accuracy you want).
Starting with x = 1 produces the sequence (4dp) 2.3964, 2.7462, 2.8007, 2.8085, 2.8096, 2.8098, 2.8098,.... .
Newton-Raphson will get you there more quickly.
Carry more decimal places if you wish.
Try a starting value of x = 2, or x = 3.
