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
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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.
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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
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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.
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Bob Krueger
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4 years, 2 months ago

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

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@Shubham Srivastava
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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 \((-2)^2\) is a real number, \((-2)^{(\frac{1}{2})}\) is not a real number.
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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.
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Harsa Mitra
·
4 years, 2 months ago

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TopNewestwe 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-adiclogarithms – Nicolae Sapoval · 4 years, 2 months agoLog in to reply

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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– Shubham Srivastava · 4 years, 2 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.Log in to reply

– Bob Krueger · 4 years, 2 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.Log in to reply

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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