normal logarithm imaginary number problem

Hello,

i have a problem. If i want to calculate the normal logarithm with an exponent.

 (log(0.5))^(1/1.55)

or

(-0.6931)^(1/1.55)

matlab results:

-0.3477 + 0.7087i

but I want an other result. If I write the equation on this way:

-0.6931^(1/1.55)

matlab results

-0.7894

Why is there a difference? The normal logarithm of 0.5 is -0.6931 and with an exponent of (1/1.55) my calculator show 0.7894 (not -0,7894 or -0.3477 + 0.7087i).

I need this (x=0.5)

(log(x))^(1/1.55)= 0.7894

Answer 1

[Stephen23]

"Why is there a difference": because the order of operations is different. In particular, the ^ has a higher priority over the -, so the calculations are actually different. With the brackets these are all equivalent:

(log(0.5))^(1/1.55)
 = (-0.6931)^(1/1.55)
 = (-0.6931)^0.6452
 = -0.3477 + 0.7087i

Whereas according to standard order of operations rules, without the brackets the following are equivalent:

-0.6931^(1/1.55)
 = -(0.6931^(1/1.55))
 = -(0.6931^0.6452)
 = -(0.7894)
 = -0.7894

This is correct and documented behavior.

If you want to get 0.7894, then you can simply take the absolute value:

>> abs(log(0.5))^(1/1.55)
ans = 0.7894

And if you need to keep the sign as well:

>> x = 0.5;
>> sign(log(x)) * abs(log(x))^(1/1.55)
ans = -0.7894