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Logarithm conceptual problem

\(f(x)=\log_a x\) is defined only for \(a>0\) and \(a \neq 1\). Why is it so? Why is it not defined for negative values of a?

Note by Shubham Srivastava
4 years, 2 months ago

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we can also get this from its graph..... Ajay Kotwal · 3 years, 2 months ago

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because if a=1 it has many possibilities Pelajar Mipa · 4 years, 2 months ago

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Does that \( \log_{-2} 4 \) exist? Well, you would believe it is equal to 2. But there's another result. \((-2)^{\log(4)/(\log(2)+i \pi)} = 4\).

That is to say, the logarithm is not uniquely defined at negative values. George Williams · 4 years, 2 months ago

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Frankly speaking you can extend logarithm notion further, for example one defines complex logarithm.

The other cases may be indeces (logarithm anologue for \(\mathbb{Z}\setminus p\mathbb{Z}\)) or p-adic logarithms Nicolae Sapoval · 4 years, 2 months ago

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The logarithm is defined as the inverse function of the exponential. Thus, \(log_a b =c \Rightarrow a^c=b\). Since the right hand side will not define b for all values of c when a is negative, it is common to restrict a to the positive numbers. It is also common to not let a=1, for then b must equal one. However, the definition of a logarithm can be extended to negative a if we allow b to be complex. In this case, new and unusual rules and situations arise. In fact, I see nothing wrong with letting a be complex too. This is a perfect example of a function that can be extended, but is commonly restricted for more practical use. Bob Krueger · 4 years, 2 months ago

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@Bob Krueger "Since the right hand side will not define b for all values of c when a is negative.", But \((-2)^2=4\) is true. So b is defined, for c=2, when a is negative. Shubham Srivastava · 4 years, 2 months ago

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@Shubham Srivastava Yes, for that particular value of c, but not for all values of c. And at this point I am still assuming that a, b, and c are real numbers. So while \((-2)^2\) is a real number, \((-2)^{(\frac{1}{2})}\) is not a real number. Bob Krueger · 4 years, 2 months ago

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a^y=x Now a can only be zero if,x=0 and a can be 1 only when y=0 So for inclusion of these values,one needs to modify the domain and range respectively. Harsa Mitra · 4 years, 2 months ago

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