Logarithm conceptual problem

f(x)=logaxf(x)=\log_a x is defined only for a>0a>0 and a1a \neq 1. Why is it so? Why is it not defined for negative values of a?

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Note by Shubham Srivastava
8 years ago

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Comments

The logarithm is defined as the inverse function of the exponential. Thus, logab=cac=blog_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 - 8 years ago

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"Since the right hand side will not define b for all values of c when a is negative.", But (2)2=4(-2)^2=4 is true. So b is defined, for c=2, when a is negative.

Shubham Srivastava - 8 years ago

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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(-2)^2 is a real number, (2)(12)(-2)^{(\frac{1}{2})} is not a real number.

Bob Krueger - 8 years ago

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

The other cases may be indeces (logarithm anologue for ZpZ\mathbb{Z}\setminus p\mathbb{Z}) or p-adic logarithms

Nicolae Sapoval - 8 years ago

Does that log24 \log_{-2} 4 exist? Well, you would believe it is equal to 2. But there's another result. (2)log(4)/(log(2)+iπ)=4(-2)^{\log(4)/(\log(2)+i \pi)} = 4.

That is to say, the logarithm is not uniquely defined at negative values.

George Williams - 8 years ago

because if a=1 it has many possibilities

pelajar mipa - 8 years ago

we can also get this from its graph.....

Ajay Kotwal - 7 years ago

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 - 8 years ago
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