When does an anonymous function reach a specified value?

I get the feeling this a rather simple question, but I cannot seem to get this to work how I want.

Consider an equation, say, y = e^(a*x). I want to find the first x-value where y crosses a threshold, say, y>100. The function is written in MATLAB as an anonymous function, i.e.,

y = @(x) exp(a*x);

How would I go about solving for x? Ideally I want something more specific/robust then simply plotting the function and eyeballing/using the plot tools to estimate the x-value.

Answer 1

[Torsten]

threshold = 100;
a = 0.5;
fun = @(x) exp(a*x) - threshold;
solx = fsolve(fun,1)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

solx = 9.2103

exp(a*solx)

ans = 100.0000

[Nicholas Keen]

Ugh, thank you. Feels silly seeing simple solutions to a simple problem.

Answer 2

[John D'Errico]

Since you do not provide a value for a, there is no numerical solution possible.

However, as long as the function is a simple one like this, the answer is just as simple. I'll use the symbolic toolbox. We just need to solve for the value of x.

syms x a
ytarget = 100;

yfun = exp(a*x);

xtarget = solve(yfun == ytarget,x)

xtarget = 

And that gives us a nice EXACT analytical result. For any value of a, we can know the value of that would achieve the desired target value for y. We can even turn this into a function we can use.

xfun = matlabFunction(xtarget)

xfun = function_handle with value:
    @(a)(log(1.0e+1).*2.0)./a

For example, when a==0.5, the answer will be...

format long g
xfun(0.5)

ans = 
          9.21034037197618

Easy enough. Of course, this will not always be the case. Then you might need to resort to numerical methods.

Answer 3

[Dyuman Joshi]

It's not possible to determine the exact value which satisfies the condition you have stated, but you can get an approximate value, by finding the minimum value of the function, which will depend upon the tolerance you define -

Using fminsearch with tolerance of 10^-10.

%Random value
a=2;

val=100;
%Original function
y1 = @(x) exp(a*x);

%Function to be minimised
y2 = @(x) abs(y1(x)-val);

%Starting point
x0 = 0;

%Set options as required
options = optimset('Display','iter','TolX',1e-10);

x = fminsearch(y2,x0,options)

 
 Iteration   Func-count         f(x)         Procedure
     0            1               99         
     1            2          98.9995         initial simplex
     2            4          98.9985         expand
     3            6          98.9965         expand
     4            8          98.9925         expand
     5           10          98.9844         expand
     6           12           98.968         expand
     7           14          98.9344         expand
     8           16           98.864         expand
     9           18          98.7089         expand
    10           20          98.3322         expand
    11           22          97.2171         expand
    12           24          92.2515         expand
    13           26          39.9306         expand
    14           28          39.9306         contract inside
    15           30         0.233253         contract outside
    16           32         0.233253         contract inside
    17           34         0.233253         contract inside
    18           36         0.233253         contract inside
    19           38         0.233253         contract inside
    20           40         0.233253         contract inside
    21           42         0.233253         contract inside
    22           44         0.166879         contract inside
    23           46        0.0329868         contract inside
    24           48        0.0329868         contract inside
    25           50        0.0170172         contract inside
    26           52       0.00798172         contract inside
    27           54        0.0045185         contract inside
    28           56       0.00173142         contract inside
    29           58       0.00139359         contract inside
    30           60      0.000168901         contract inside
    31           62      0.000168901         contract inside
    32           64      0.000168901         contract inside
    33           66      2.64113e-05         contract inside
    34           68      2.64113e-05         contract inside
    35           70      2.24168e-05         contract inside
    36           72      1.99725e-06         contract inside
    37           74      1.99725e-06         contract inside
    38           76      1.99725e-06         contract inside
    39           78      1.05451e-06         contract inside
    40           80      4.71367e-07         contract inside
    41           82      2.91572e-07         contract inside
    42           84      8.98975e-08         contract inside
    43           86      8.98975e-08         contract inside
    44           88      5.46999e-09         contract inside
    45           90      5.46999e-09         contract inside
    46           92      5.46999e-09         contract inside
    47           94      5.46999e-09         contract inside
 
Optimization terminated:
 the current x satisfies the termination criteria using OPTIONS.TolX of 1.000000e-10 
 and F(X) satisfies the convergence criteria using OPTIONS.TolFun of 1.000000e-04 

x = 2.3026

format long
%x value
x

x = 
   2.302585093021396

%Value of the original function at the point
y1(x)

ans = 
     1.000000000054700e+02

%Value of the function that was minimised at the point
y2(x)

ans = 
     5.469985353556694e-09

'TolX' is the termination tolerance for x. Read the documentation page of fminsearch for more info.