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What is the domain of $f(x)=x^{2x}$?

If $f(x)=(x^2)^x $then $f$ is defined for every real number but if $f(x)=(x^x)^2$ then $f$ is only defined and "nice" (excluding the negative $-p/q$ fractions) for positive real numbers.

Should we say $f(x)=e^{2x\log(x)}$ is only defined for positive $x$?

Thanks

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  • $\begingroup$ For $f(x)=(x^x)^2$, the function is defined for all real numbers except 0. And for $f(x) = e^{2xlog(x)}$, the domain is only positive numbers. $\endgroup$ – harshit54 Nov 17 '18 at 13:36
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    $\begingroup$ @harshit54 Really? What is $f(x)=(x^x)^2$ for $x=-\frac14$? $\endgroup$ – Servaes Nov 17 '18 at 13:38
  • $\begingroup$ @Servaes Okay, sorry. So it's defined for all positive reals, and negative integers. $\endgroup$ – harshit54 Nov 17 '18 at 14:02
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I would say that it's defined only for positive numbers. Let's look at a simpler problem: what is the domain of $x^\frac12$? I can say "I could always write it as $(x^2)^\frac14$." The issue is order of operations. Unless you have parantheses, you need to calculate the exponent first. See for example https://en.wikipedia.org/wiki/Order_of_operations#Serial_exponentiation

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  • $\begingroup$ Nice example! Thanks $\endgroup$ – Pedro Nov 17 '18 at 15:44

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