Determining logarithm's linearity

[atrus_ovis]

Is the following sufficient to prove linearity of
y(t) = log (x(t)) ?

In order for the system to be linear, this must be true:
ax₁(t) + ax₂(t) $$\stackrel{response}{\rightarrow}$$ a log(x₁(t)) + a log(x₂(t) ) (1)
but for input ax(t):
ax(t) -> log(ax(t))=log(a) + log(x(t))

thus
ax₁(t) + bx₂(t) $$\stackrel{response}{\rightarrow}$$ log(ax₁(t)) + log(bx₂(t) )
=log(a) + log(x₁(t)) + log(b) + log(x₂(t)) =/= (1)

is this sufficient?

i don't know if this counts as a homework question, it's kinda general.Sorry if I'm wrong.

 

[mathman]

ax1(t) + bx2(t) = a log(x1(t)) + b log(x2(t) ) (1)

Where did you get this?

 

[atrus_ovis]

Ah, wrong.The constant must be the same. Also, no '=', i used it as 'the system yields' operator.

edited

 

[vela]

Ah, wrong.The constant must be the same.

Actually, the constants don't have to be the same. To show a function f is linear, you need to show two things: 1) f(x+y) = f(x)+f(y) and 2) f(ax) = a f(x). You can show both simultaneously by proving f(ax+by) = a f(x) + b f(y).

Also, no '=', i used it as 'the system yields' operator.

This is a really bad practice, which should be avoided and which, unfortunately, students commonly use. In mathematics especially, the equal sign is never used this way.

 

[vela]

ax₁(t) + bx₂(t) = log(ax₁(t)) + log(bx₂(t) )

How did you get this?

log (ax + by) ≠ log(ax) + log(by)

is this sufficient?

What are you trying to show specifically?

 

[atrus_ovis]

damn, the systems notation confused me

What are you trying to show specifically?

to determine if the system x(t) ->log(x(t)) is linear.Meaning that the system which at an input x(t) produces an output of log(x(t)) is linear.

so i have to show that
it's homogenous
log (ax(t)) = a log(x(t))
and additive
log(x₁(t) + x₂(t)) = log(x₁(t)) + log( x₂(t))
right?

by the way, does proving the above mean / is the same as saying that it's closed under scalar multiplication and closed under addition?

 

[vela]

damn, the systems notation confused me


to determine if the system x(t) ->log(x(t)) is linear.Meaning that the system which at an input x(t) produces an output of log(x(t)) is linear.

so i have to show that
it's homogenous
log (ax(t)) = a log(x(t))
and additive
log(x₁(t) + x₂(t)) = log(x₁(t)) + log( x₂(t))
right?

Right. Or show that either one doesn't hold and conclude that the system isn't linear.

by the way, does proving the above mean / is the same as saying that it's closed under scalar multiplication and closed under addition?

It looks similar, but it's not the same. When you talk about closure, you're typically referring to a set and an operation. You'd say the set is closed under the operation if applying that operation to any two elements in the set yields another element in the set. For example, you'd say the set of integers Z is closed under addition because if you add any two integers, you'll get another integer. This same set would not be closed under division because while 1 and 2 are both in Z, their quotient, 1/2, is not an element of Z.

 

[atrus_ovis]

So will .. simply mentioning that

log(ax(t)) = loga + log(x(t)) =/= alog(x(t)) ,

log(x₁(t)+x₂(t)) =/= log(x₁(t)) + log(x₂(t)),

due to the logarithm's properties.Thus the system y(t)=log(x(t)) is neither homogenous nor additive, and thus it is not linear.

will suffice?

 

[vela]

Yes, simply noting either one would be sufficient to show the log isn't a linear function.

 

[atrus_ovis]

All right.
Thanks for the helpful replies.

 

[HallsofIvy]

log 1+ log 1= 0+ 0= 0 while log 2 is NOT 0. That suffices.

 

[atrus_ovis]

Great. Thanks.