Waste less time on Facebook — follow Brilliant.
×

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, 7 months ago

No vote yet
3 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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

Ajay Kotwal - 3 years, 7 months ago

Log in to reply

because if a=1 it has many possibilities

Pelajar Mipa - 4 years, 7 months ago

Log in to reply

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, 7 months ago

Log in to reply

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, 7 months ago

Log 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, 7 months ago

Log in to reply

"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, 7 months ago

Log in to reply

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, 7 months ago

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, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...