How can calculate the not scalar in Mablab?

Question

I'm looking for a method about the power calculation in Matlab. The code like s following

Corrected = 255 * (Image/255)^(1/2.2)

but as you can see this, this make an error like this:

Error using ==> mpower
error happened like this. Integer can only be raised to positive integra powers

So what can I do fix it?

@Divakar Yep, the image values are not integer the real numbers. So I want to know another ways — Carter, Mar 04 '15 at 06:26
you forgot about `/255`. `Image` is not a scalar. So all operators should be array (with `.` preceding). — Kirill Slatin, Mar 04 '15 at 06:28
@qazi Convert `image` to `double()` and then use the equation? — Divakar, Mar 04 '15 at 07:02

score 0 · Answer 1 · answered Mar 04 '15 at 06:22

0

Im not completely sure what you are asking but, try using the . operator so do .^ instead of ^

answered Mar 04 '15 at 06:22

David

Also error happened like this. Integer can only be raised to positive integra powers. – Carter Mar 04 '15 at 06:29
use combination of `log` and `exp` – Kirill Slatin Mar 04 '15 at 06:31
@KirillSlatin How can I fix it? – Carter Mar 04 '15 at 06:37
mmm how do you use matlab if you don't know basic math? In your case should be something like `Corrected = 255 .* exp((1/2.2) .* log(Image ./ 255))` – Kirill Slatin Mar 04 '15 at 06:50
[http://en.wikipedia.org/wiki/Exponentiation#Powers_via_logarithms](http://en.wikipedia.org/wiki/Exponentiation#Powers_via_logarithms) – Kirill Slatin Mar 04 '15 at 06:56
@KirillSlatin thanks but it still have error like this. : undefined function or method 'log' for input argument of type 'uint8'. – Carter Mar 04 '15 at 07:05
Check [here](http://stackoverflow.com/questions/14904030/undefined-function-log-for-input-arguments-of-type-uint8). You need to convert your Image using `double()` – Kirill Slatin Mar 04 '15 at 07:08

score 0 · Accepted Answer · edited May 23 '17 at 11:43

0

Corrected = 255 .* exp((1/2.2) .* log(double(Image./255)))

edited May 23 '17 at 11:43

Community

answered Mar 04 '15 at 06:28

Kirill Slatin

Yep, I found the way as your recommend, I would have to do like this. log(double(Image)) – Carter Mar 04 '15 at 07:24
Yeah, thanks, I corrected the original answer once again – Kirill Slatin Mar 04 '15 at 07:26
Hi both of them equations are totally different. – Carter Mar 05 '15 at 07:13
I mean that your equation results and Use Corrected = 255 * (Image/255).^(1/2.2) are different. – Carter Mar 05 '15 at 09:30
I mean that I need to approximation methods instead of using log or exp functions. – Carter Mar 05 '15 at 09:32
Seems to be completely out of scope of question initially asked. You mean you are forced to solve the problem using approximation methods? – Kirill Slatin Mar 05 '15 at 09:48
Yes Slatin, do you know what approximation methods? I have just made new question regarding that but I have down vote 2 already . – Carter Mar 05 '15 at 11:01

2 Answers2