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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, 4 months ago

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

- 3 years, 4 months ago

because if a=1 it has many possibilities

- 4 years, 4 months ago

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.

- 4 years, 4 months ago

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

- 4 years, 4 months ago

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.

- 4 years, 4 months ago

"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.

- 4 years, 4 months ago

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.

- 4 years, 4 months 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.

- 4 years, 4 months ago