Is this a valid proof?

[nacho]

 where is false by definition







Proof by contradiction
/end

[taiga]

no

[taiga]

Well I suspect you are not finished with your post yet so I will edit later if it starts to make more sense to me

[BubbleWrapMan]

I'm not sure what you're trying to prove.

Logarithms with negative bases aren't something VCE deals with as far as I know.

[nacho]

The proof is showing that you cannot take the log of negatives
I was confused though,
why can't you take the log of a negative number, given that the base is a negative number?

ex.



My friend said that it would not yield a smoothe curve.
But if i were to answer the question correctly, would i put "Undefined" or "3" ?

[brightsky]

You technically can have a negative base, but the log function would be very 'retarded'.
Let's define log_a(x) = y, where a < 0. Then a^y = x
When the base is negative, the log would only be defined for when y is an integer, since (-something)^integer is gives -something^3. But when y = 1/(2n), for n E Z, then we have some problems. Consider, say (-2)^(1/2). It's undefined. Consider (-2)^(1/4), it's undefined. Hence if we WERE to sketch the graph, it will be a very screwed up one. Some people say that because not all y yield a defined log function, then the log isn't really defined.

[luffy]

You technically can have a negative base, but the log function would be very 'retarded'.
Let's define log_a(x) = y, where a < 0. Then a^y = x
When the base is negative, the log would only be defined for when y is an integer, since (-something)^integer is gives -something^3. But when y = 1/(2n), for n E Z, then we have some problems. Consider, say (-2)^(1/2). It's undefined. Consider (-2)^(1/4), it's undefined. Hence if we WERE to sketch the graph, it will be a very screwed up one. Some people say that because not all y yield a defined log function, then the log isn't really defined.

Actually, the log function is not defined for when the base is negative, even when you might think it would yield a defined result. It is only valid over the complex field.

This is because a logarithm of any given base is actually defined to be under the natural logarithm "e" using the change of base law. Below is an example:
-> It is defined such that
Therefore, for any negative base, calculations are done using complex numbers:






Therefore, the log function of a negative base does not exist for ANY real values. It only works in the complex plane.

Just thought I would mention it in the methods board

[nacho]

You technically can have a negative base, but the log function would be very 'retarded'.
Let's define log_a(x) = y, where a < 0. Then a^y = x
When the base is negative, the log would only be defined for when y is an integer, since (-something)^integer is gives -something^3. But when y = 1/(2n), for n E Z, then we have some problems. Consider, say (-2)^(1/2). It's undefined. Consider (-2)^(1/4), it's undefined. Hence if we WERE to sketch the graph, it will be a very screwed up one. Some people say that because not all y yield a defined log function, then the log isn't really defined.

Actually, the log function is not defined for when the base is negative, even when you might think it would yield a defined result. It is only valid over the complex field.

This is because a logarithm of any given base base is actually defined to be under the natural logarithm "e" using the change of base law. Below is an example:
-> It is defined such that
Therefore, for any negative base, calculations are done using complex numbers:






Therefore, the log function of a negative base does not exist for ANY real values. It only works in the complex plane.

Just thought I would mention it in the methods board

nice work nags

[dooodyo]

how does e^(loge(5)) = 5 ?

[Water]

e^(loge(5)) = 5



General Rule:


f(x) composite f-1(x) = x

In other words.
f(f-1(x)) = x   

[Souljette_93]

how does e^(loge(5)) = 5 ?

They're the inverse of each other, so they cancel out i.e.  x * 1/x =1

[dooodyo]

Oh haha thanks but how would you prove it?

[nacho]

how does e^(loge(5)) = 5 ?

let

then

hopefully, VCAA finds that to be an acceptable proof

[Water]

Basic equations to prove it.


y = 2x

Inverse

y= x/2




Composite of f(f-1) = 2(x/2)
                           = x


Composite function f(f-1)  , if x = 1  

           2(1/2) = 1

If x = 2

  2(2/2) = 2






Normal Function


2^x


Inverse Function

log2 (x) = y  


f(f-1) = 2^log2(x)


If the composite function had x = 2

                       2^log2(2) = 2^1 = 2

If the composite function had x = 4

                  2^log2(4)

                  2 ^ 2 = 4

Therefore 2^log2(x) = x


Thereby means.

Normal Function, Composite with Its Inverse = x

[brightsky]

let

then

hopefully, VCAA finds that to be an acceptable proof

Yep, this is the most straightforward one.