Inverse of log formula

Hi, I have the following formula:

    y = -log(x / mean(x))

y and x are vector’s.

However, in my case I only know y, but not x. Therefore, I would like to solve the inverse of the formula for x. I am aware of that exp is the inverse of log:

    x=-exp(y)

but I am not sure how to consider the term mean(x)?

Thank you in advance.

Lisa

Answer 1

[Star Strider]

You cannot determine ‘mean(x)’ unless you have access to the original ‘x’ vector. (If you had ‘x’, you would not need to do the inversion.) You can only recover the normalised ratio ‘x/mean(x)’.

[John D'Errico]

Star is exactly correct here. The scaling by the mean of x essentially loses that information forever. It cannot be recovered, knowing only y.

[Lisa]

Thank you very much.

[Star Strider]

@Lisa — As always, my pleasure!

@John D’Errico — Thank you!

Answer 2

[Image Analyst]

Try fzero().

[John D'Errico]

Nope. Not possible to use fzero here.

Answer 3

[Image Analyst]

Lisa, is this what you are after - to find the x for a y that is a specified value? You can do it numerically like this, where I find the x where y = 1:

% Define a range of x values.
x = sort(rand(1, 500), 'ascend');
% Compute the y values over that range.
fprintf('mean(x) = %f\n', mean(x));
y = -log(x / mean(x));
% Plot curve.
plot(x, y, 'b-', 'LineWidth', 2);
grid on;
xlabel('x', 'FontSize', 20);
ylabel('y', 'FontSize', 20);

% Find the x closest to y = 1 (for example).
[minValue, index] = min(abs(y-1))
% Get x and y values
x1 = x(index);
y1 = y(index);
fprintf('Closest point: y = %f at x = %f\n', y1, x1);
% Draw lines
hold on;
line([0, x1], [y1, y1], 'Color', 'r', 'LineWidth', 2);
line([x1, x1], [0, y1], 'Color', 'r', 'LineWidth', 2);
ax = gca;
ax.XAxisLocation = 'origin';

In the command window:

mean(x) = 0.514679
minValue =
        0.0017406261319084
index =
    93
Closest point: y = 1.001741 at x = 0.189011

And the plot of the results: