Ok, so I have a problem that basically boils down to whether or not ln(-6) is a thing. My teacher said it isn't because of the rule that the argument can't be negative. My calculator on the other hand says ln(-6)=1.791759469+πi. And if you rearrange it is in fact true that e^1.791759469+πi is -6. Can someone please explain why I'm wrong?
Thanks!
Although logarithms are usually undefined for arguments of 0 and negative values, it seems that your calculator and rearrangement is not completely wrong and undefined (though without imaginary and complex numbers it will be undefinable). First, ln(6) does have decimal approximation of 1.791759469. Next, e^(pi i) equals -1. (See the link at the end for a site outside of Khan Academy that explains why.)
If you meant that the entire "1.791759469+πi" is actually the exponent for e (instead of what will result in 6+πi if πi was added after you got 6), the answer is -6+0i (or just -6). When we multiply values that contain exponents, when the bases are the same we can keep the base and add the values of the exponents together (e.g (2^3)*(2^4) can be rewritten as 2^7). By applying the reverse rule of how we can add exponents together, we can split the e^(1.791759469+πi) into e^(1.791759469) * e^(πi). e^1.791759469 will give you 6, and e^(πi) will give you -1. Multiplying these two values together gives -6, so viewing it that way means it is a correct answer.
While logarithms may be taught at Algebra 2, complex numbers may not be taught until Precalculus or a class/course after Algebra 2.
https://www.math.toronto.edu/mathnet/questionCorner/epii.html
A bejegyzéshez nem írhatók újabb hozzászólások.