+33657 It is also possible that the equation in question is meant to be:
e−0.2x−e−0.75x=0.25
In this case there are two solutions as can be seen from the graph below:
These solutions can be obtained numerically. A simple iterative approach is to write the equation in the following form:
xn+1=−10.2ln(0.25+e−0.75xn)
Then let x0 = 1, say, and use the formula above to iterate several times. The result is:
x ≈ 6.812073
By rewriting the iteration formula as:
xn+1=−10.75ln(e−0.2xn−0.25)
we can iterate to the other solution:
x ≈ 0.6023554654
+118703 How about with the site calculator?
e(−0.2)×0.25−e(−0.75)×0.25=0.0865910500842418
I am not sure if this is actually what you meant - you must be careful to check the brackets are in the right place.
(It is what you asked i believe)
+130477 Maybe you meant this:
e^(-.2) x - e^(-.75) x = .25 ??
I'm going to assume so......So we have
0.8187307530779818 x - 0.4723665527410147 x = .25
0.3463642003369671 x = .25 So x = .25 / 0.3463642003369671 = 0.7217836016446927047
+118703
Chris thinks you meant
(e−0.2×x)−(e−0.75×x)=0.25
Yes, this probably is what you meant. Your title is missing a bracket so there is plenty of room for confusion!
I read in an extra x (without even realising) and Chris read in an extra bracket. I now see your main text does have the bracket!
+33657 It is also possible that the equation in question is meant to be:
e−0.2x−e−0.75x=0.25
In this case there are two solutions as can be seen from the graph below:
These solutions can be obtained numerically. A simple iterative approach is to write the equation in the following form:
xn+1=−10.2ln(0.25+e−0.75xn)
Then let x0 = 1, say, and use the formula above to iterate several times. The result is:
x ≈ 6.812073
By rewriting the iteration formula as:
xn+1=−10.75ln(e−0.2xn−0.25)
we can iterate to the other solution:
x ≈ 0.6023554654
