[ e^(-1i*a*x) ]^(1i / a*x) - e = 0 , substitute,a=2, solve,x . How many solutions ?

[lvl107]

Hello,

Thanks in advance.

[LucMeekes]

The answer should be:infinite

(e^B)^C=e^(B*C)

in your case B=-iax and C=i/(ax), so B*C=1

That simplifies your equation to e-e=0

any value of x will do, as long as it's unqual to 0.

Success!

Luc

[Werner_E]

(e^B)^C=e^(B*C)

As we are obviously dealing with complex numbers, thats only true if C is an integer! Otherwise any complex number with magnitude e^(B*C) and argument any integer multiple of C*2*pi is result. If C is irrational thats even an infinte number of results.

But as the question is about solving that equation, your solution is valid nonetheless: if a is unequal to zero, x can be any value different from zero, otherwise x is undefined.