complex function inverse plot

Hello, i have a plot a function

y=1/(ln(x)-0.1)

how can i plot X as a function of Y ,when y=[5:0.01:6]

Thanks.

[Ankit]

syms x
y=(5:0.01:6);
y=1./(log(x)-0.1);
g = finverse(y);
plot(y,g);

For this you need symbolic tool box, I would like you to read this link: https://de.mathworks.com/matlabcentral/answers/286749-find-inverse-of-a-function

[fima v]

Hello Ankit, When i ran your code it gave m an error on the plot command

"DATA must be numeric,datetime,duration or an array convertible to double"

g,y has dimentions of 1X1

where is the problem?

Thanks.

Answer 1

[KSSV]

On solving it manually....we have:

y = 1/(log(x)-0.1) ;
x = exp(1/y+0.1) ; 

To plot:

y=[5:0.01:6] ;
x = exp(1./y+0.1) ; 
plot(x,y)

Answer 2

[John D'Errico]

Note that the natural log function is NOT written as ln in MATLAB, but just as log.

Next, in general, it may be impossible to plot your relationship, because there may be infinitely many disjoint regions over which your plot would need to be generated. For example, suppose you wanted to plot the regions where something as simple as sin(1./x) is between .5 and 1?

If a solution exists, where the function has a simple inverse, and if you have the symbolic toolbox, then finverse can help you.

syms X
Y = 1/(log(X) - 0.1)
fplot(finverse(Y),[5,6]);
xlabel 'Y'
ylabel 'X'
grid on

Or, if you want to use plot, you could do it like this:

syms X
Y = 1/(log(X) - 0.1);
y = 5:.01:6;
xfun = matlabFunction(finverse(Y));
plot(y,xfun(y),'-');
xlabel 'Y'
ylabel 'X'
grid on

If you don't have the symbolic toolbox, then you would need to use a loop, and then use fzero to compute the inverse.