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If we accept both positive and negative values for the square root of a number, then can the anti-log of a number be negative?

Yashbhatt
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    It can be negative if you consider the complex field like you do with square root of a negative number. – lsp May 13 '14 at 05:49
  • But the square root of a negative number is not a real number whereas anti-log gives us real values. – Yashbhatt May 13 '14 at 05:50
  • You are allowed to use the traditional name, "exponential function". The exponential function has only positive values over the reals. – Lutz Lehmann May 13 '14 at 06:48
  • And please define what you think the logarithm is, from most of the motivations, historical and recent, it is clear that the (real-valued) logarithm is only defined for positive numbers. – Lutz Lehmann May 13 '14 at 07:27
  • A logarithm is actually the power to which the base needs to be raised to get a particular number. – Yashbhatt May 13 '14 at 14:13
  • I've read in some other threads that we always take positive values for square root. Why is that? – Yashbhatt May 13 '14 at 14:16

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can the anti-log of a number be negative?

If by “anti-log” you mean the exponential function, then we have Euler's famous identity $e^{i\pi}=-1$

Lucian
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  • By anti-log I mean what we find in anti-log tables. I don't know if its called exponential function. – Yashbhatt May 13 '14 at 16:54
  • So basically $\log x=a$, where a is known, and x is the unknown ? In this case, you simply apply the fact that $\log\dfrac1x=-\log x=-a$. – Lucian May 13 '14 at 17:03
  • I get what you have written. But what I meant by my question was that suppose you have a number $x$ such that $log x $ = $a$. Now, to find out the square root of $x$, we multiply $a$ by 1/2 and find the anti log of the obtained number. But as we know the square root of $x$ can be negative or positive , then shouldn't we also have negative value for anti-log? – Yashbhatt May 14 '14 at 13:44
  • Not negative, but complex, since $\log\Big(-\sqrt x\Big)=\log\Big((-1)\cdot\sqrt x\Big)=\log\Big(-1\Big)+\log\Big(\sqrt x\Big)=$ $=\log e^{(2k+1)\pi i}+\dfrac12\log x=\dfrac a2+(2k+1)\pi i$. – Lucian May 14 '14 at 20:14
  • But we still have to find $antilog (\dfrac{a}{2}) + antilog (2k+1)\pi i$ to get the answer and how do we find that? – Yashbhatt May 15 '14 at 04:54
  • $e^{x+y}\neq e^x+e^y$. Rather, $e^{x+y}=e^x\cdot e^y$. The number you're looking for is $-\sqrt{e^a}$. – Lucian May 15 '14 at 04:59
  • Ok. So, the it's $antilog [\dfrac a2+(2k+1)\pi i]$, right? – Yashbhatt May 15 '14 at 05:34
  • Yes. $\log\Big(-\sqrt{e^a}\Big)=\dfrac a2+(2k+1)\pi i$. – Lucian May 15 '14 at 06:33
  • Ok. Thanks. But we don't use negative values for anti-logs in calculations. Why is that? – Yashbhatt May 15 '14 at 10:08
  • Sorry, I've just realized that I hadn't understood your question. You were talking about the case $\log_xa=b$, and what I had in mind was $\log_ax=b$. In which case the answer is simple: because a function can only have one value in any given point. Just like in the case of square roots, when we choose only the positive value from among the two possible results. $\sqrt9=3$, not $-3$, and $\log_x9=2$ returns only $x=3$, not $x=-3$. After all, it is a bit redundant to mention both, isn't it ? – Lucian May 15 '14 at 10:36
  • But in some cases it may be non-trivial to mention both. Think about it. – Yashbhatt May 16 '14 at 06:21
  • Do you have something specific in mind ? – Lucian May 16 '14 at 12:40