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Name: William
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Question:
While doing geometric sequences in class, we learn the formula: An = A1 * r^(n-1) now a question hit me one day... -64 = 2 * (-2)^(n-1) I know that the answer to this is 5, however my calculator says otherwise. Is there are reason why these complex numbers are being returned to me? Shouldn't log-2(-32) be a real number?



Replies:
William,

Most calculators solve this problem with natural logarithms. If b=a^x, then x=ln(b)/ln(a). Using complex numbers allows one to take a logarithm of a negative number.

Consider the equation y=ln(-x), where x itself is positive. From this, we can say that -x=e^y. It so happens that the exponential function of an imaginary number is e^(bi)=cos(b)+i*sin(b), provided that b is in radians. Going further: e^(a+bi)=(e^a)*[e^(bi)]=(e^a)*[cos(b)+i*sin(b)] The quantity "a" determines the size. The quantity "b" determines the sign. Since cos(pi)=-1, we get: ln(-x)=ln(x)+3.14i I expect your calculator gives, for -32=-2^n, n=[ln(32)+3.14i]/[ln(2)+3.14i].

This is true because your calculator works with real numbers rather than integers. -32=(-2)^n, and other such relationships, work only when n is an odd integer. Consider -31.9999=(-2)^n. n is neither odd nor even. What do we do with the sign? Complex numbers give us a way. When working with real number functions, you have to work out what is happening with your brain, reducing the expression to something less involved, before you give it to the calculator. A calculator does not "know" what it is doing. It sticks numbers into a preprogrammed function and then gets out the answer. It has no awareness of what the numbers are. They break down to just a sequence of ones and zeroes, nothing more.

Dr. Ken Mellendorf
Physics Instructor
Illinois Central College


You have a number of issues imbedded in your inquiry. Because of the inherent limitations of fonts and the limitations of expressing mathematical expressions available to this web site it is difficult to communicate exactly what your question is. For example the string "log-2(-32)" could mean different things if you mean (-2) x (-32) = + 64 or if you mean (-2)^(-32) i.e. (-2) raised to the power (-32), or do you mean the logarithm

base (-2), which extends the question into complex number theory. Could you re-phrase the issue in more detail so that it becomes clearer exactly what you are asking?

Vince Calder



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