Matlab solver for natural log equation

Hello,

I thought this was a simple solver equation, I know I am doing something wrong, because the result should be about S=1429.

This is my code:

syms  x 
eqn = log(0.5) == (-27109/x)-0.95*log(x)+25.18
S = vpasolve(eqn,x)

Matlab gives me a very large S =672919621689.2986708066438355574

What am I doing wrong here?

Answer 1

[John D'Errico]

Does a solution near(er) to zero exist? PLOT IT!!!!!!

syms x
f = (-27109/x)-0.95*log(x)+25.18 - log(0.5);
fplot(f,[10,1000])

Do you see anything strange? A solution would correspond to a point where that curve crosses the line y == 0. Looking a little further up, we finally see this:

fplot(f,[200,1e4])
yline(0);

So a solution seems to occur near x == 1500.

S = vpasolve(f,x,1500)

S = 

1428.9171626682572235629898202448

However, it you look further out, the curve reaches a peak, and then starts to drop again. It may well cross zero again out there.

fplot(f,[10000,1e6])
yline(0);

Did you tell vpasolve which root it should find? Should it know?

vpasolve is a numerical root finder. It looks for a root, near where you tell it to look. If you don't tell it where, it picks a spot. And sometimes, it may end up converging to a solution you don't like.

[Shane Palmer]

Thank you very helpful.

Answer 2

[Walter Roberson]

Q = @sym; %convert to rational
syms  x 
eqn = log(Q(0.5)) == (-Q(27109)/x)-Q(0.95)*log(x)+Q(25.18)

eqn = 

S = solve(eqn,x)

S = 

vpa(S)

ans = 

You can see from this that there are two solutions. You could use an initial value in vpasolve to get the other one