<p>Hi! There's this thing that's been bugging me with logarithms. If we have log_1 (1), what should it be equal to? My calculator gives it out as an error, and I have an algebra software that yields it 1. Shouldn't the answer be infinite, really? I mean 1^4 = 1 and 1^-2 = 1 also. So what's the right and most appropriate answer? i.e. infinity or all real numbers or something like that (I'm not really good at mathematical terminologies).</p>
<p>Another question that just popped in my mind, can we use imaginary numbers as exponents? My algebra software says that 1^i = 1 but says that "Further simplification is not possible" when I type 2^i for example. My calculator says that they're math errors in both cases.</p>
<p>By the way, I use a scientific calculator. Thanks in advance! :)</p>
<p>The log function is not defined for base 1 for exactly this reason: log base 1 of 1 would not be unique, and log base 1 of any other number would not exist.</p>
<p>Thanks Sikorsky for your quick reply! Can you answer the other question about the imaginary numbers too please?</p>
<p>Also, I have another question related to the previous log question: I keep getting errors in both my calculator and my algebra software when I type a negative number number in the base of the log. Ex: log_-2(4). Shouldn’t this be 2 since that (-2)^2 = 4 ? I just realized that the base of the log must be > 0 and not 1. I understand why the 1 is undefined, but can you elucidate on why the base must be > 0 ? Thanks again!</p>
<p>Somebody else will have to help you with your imaginary-number question. But I can help with your second question about logs.</p>
<p>Logarithm is defined as the inverse of the exponential function, and exponential functions are not defined for negative bases. While integer powers of negative bases are easy to do, it’s not clear what an entire graph of f(x)=(-2)^x ought to look like. Would it be continuous? Clearly not, since f(1/2) would not be real. And how would we evaluate f(pi)? It’s so thorny a question that we define exponential functions (and therefore logs) only for positive bases not equal to 1.</p>
<p>Thanks Sikorsky for your reply. I kinda got it :)</p>
<p>This is way beyond the subject tests, but here is your answer:</p>
<p>1^i = e^(ln(1^i)) = e^(i ln 1) = cos(ln 1) + i sin(ln 1) = 1 + i * 0 = 1</p>
<p>2^i = e^(ln(2^i)) = e^(i ln 2) = cos(ln 2) + i sin(ln 2)</p>
<p>(you can use your calculator to get cos(ln 2) etc.)</p>
<p>The cosine and sine come from Euler’s identity. (The number e is named after him as well.)</p>
<p>fignewton is almost correct, its actually Euler’s formula. </p>
<p>And this corresponds to complex exponentials and Taylor series, basically Calculus II material.</p>