Obtaining complex solutions to exponential equations

[cosmic trellis]

In lesson, we were asked to solve for x, where 4^(2x+1) = 8^(2x-1). The teacher only wanted the real solution, which I believe to be x= 5/2. However, I wondered if there were any other solutions, and lo and behold, wolfram alpha comes up with infinitely many. How do you solve for these solutions, & why are there infinitely many? How can you generally solve for all solutions of an exponential equation?
Thank you in advance.

[cosmic trellis]

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could some1 just say hi so I know ppl can see this post? I'm getting a red error under my last message

[modern phoenix]

since we can define exponentiation as: as a^b = e^(b ln(a)) and use euler's formula: e^ia = cos(a) + isin(a), for a given complex solution x = a+bi, we can rearrange the equation a bit:
4^(2a+1+2bi) = 8^(2a-1+2bi)
e^(2a ln(4) + ln(4) + 2bi ln(4)) = e^(2a ln(8) - ln(8) + 2bi ln(8))
4e^(2a ln(4)) e^(2bi ln(4)) = 1/8 e^(2a ln(8)) e^(2bi ln(8))
now just use euler's formula and you'll get:
4(cos(2b ln(4))e^(2a ln(4)) + isin(2b ln(4))e^(2a ln(4)))= 1/8(cos(2b ln(8))e^(2a ln(8)) + isin(2b ln(8))e^(2a ln(8))
since sin and cosine are periodic, its pretty easy to show that there are infinitly many solutions for this problem.
this is only a guess tho, i'm not 100% sure if this makes sense, so if anyone can correct me i'd be glad ^^

[cosmic trellis]

Thanks

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