Equaton problem

Of course, 5n(2i)^n-1 is the same as 5n(2^n-1)(i^n-1). But i to any power is one of i, -i, -1, and 1. Since the right hand side of the equation is a positive real number, i^n-1 must be equal to 1. This equation is exactly the same as n (2^n-1)= 11520 (with the provision that n-1 must be a multiple of 4 so that the power of i is, in fact, +1). You still are not going to be able to solve that equation using logarithms- you have n both inside and outside a logarithm.

But since n must be a positive integer, there is nothing wrong with "trial and error"- it's just a little tedious. If n=1, then n-1= 0= 4(0), but obviously 5(1)(2⁰)= 5 is NOT 11520. Okay, try n-1= 4, the next multiple of 4, so n= 5. Now 5(5)(2⁴)= 25(16)= 400, not 11520. The next multiple of 4 is 8:if n-1= 8, then n= 9. 5(9)(2^{8)= 45(256)= 11520. Got it!}

 

I agree with HOI, trial and error is the way to go. But, you can severely limit the possible values of n by simply looking at the original statement.

[tex] 5n \cdot (2i)^{n-1} = 11520 [/tex]

Now, if we assume n > 0 then 5n > 0, and therefore, so must the exponential. But, this will only occur for multiples of 4 plus 1. In other words:

[tex] n = 4m + 1 [/tex] for some m = 0, 1, 2,...

A quick check of these values will land you an answer at m = 2, and thus, n = 9. Since the absolute value of the expression on the left is increasing without bound for n > 0, this is the only positive solution.

If, on the other hand, we assume n < 0, then 5n < 0, and therefore, so must the exponential. This will only occur for multiples of 4 minus 1. Thus:

[tex] n = 4m - 1 [/tex] for some m = 0, -1, -2,...

However, note that for m = 0, we get a value of 5/4, and that for larger values of m, this result must be smaller. Therefore, there are no negative solutions.