Is the answer correct??

[Darya]

We have a function such that [MATH]f(x+y)=f(x*y)[/MATH][MATH]f(1)=3[/MATH]What's [MATH]f(2)[/MATH]?

It seems like it's a function [MATH]3^x[/MATH] but I'm not sure. Then the answer would be 2. Is it correct?

 

[firemath]

Do you mean [MATH]f(x+y)=f(x*y)=f(1)=3[/MATH]? I'm having trouble understanding the notation of this one also.

 

[Darya]

Do you mean [MATH]f(x+y)=f(x*y)=f(1)=3[/MATH]?

Hi! No, it's [MATH]f(x+y)=f(x∗y)[/MATH]and [MATH]f(1)=3[/MATH]

 

[firemath]

Aha! I see.

 

[JeffM]

We have a function such that [MATH]f(x+y)=f(x*y)[/MATH] [MATH]f(1)=3[/MATH]What's [MATH]f(2)[/MATH]?

It seems like it's a function [MATH]3^x[/MATH] but I'm not sure. Then the answer would be 2. Is it correct?

I am sorry but this question makes absolutely no sense.

You start with f(x + y), which implies that f has TWO arguments. Then you talk about f(1) and f(2), each of which has ONE argument.

You do not give the domain of x and y. I am going to guess that f is defined over all ordered pairs of real numbers.

 

[Darya]

I am sorry but this question makes absolutely no sense.

You start with f(x + y), which implies that f has TWO arguments. Then you talk about f(1) and f(2), each of which has ONE argument.

You do not give the domain of x and y. I am going to guess that f is defined over all ordered pairs of real numbers.

Thank you! Then that means it's a wonderful test I've found.
I guess It would make sense if [MATH]f(x+y)=f(x)*f(y)[/MATH]?

 

[firemath]

As you can see, the graph has the correct answer for [MATH]f(x)=3^x[/MATH].

But I also don't see the point of the first equation. Where is f(y)? *What* is f(y)? Do we need it? Is that what you are trying to find out?

At first I thought that [MATH]x=y=2[/MATH], but then we had conflicting arguments given to us in the notation.

See! The question is so silly that my answer telling you it is silly sounds ridiculous.

 

[firemath]

Thank you! Then that means it's a wonderful test I've found.
I guess It would make sense if [MATH]f(x+y)=f(x)*f(y)[/MATH]?

I don't know.
This is the graph of [MATH]f(x)=3^x[/MATH] and [MATH]f(x)*f(y) f(x+y)[/MATH].

 

[JeffM]

If these problems came from a text book, please give the complete wording exactly as given in the book.

 

[JeffM]

There are functions such that

[MATH]f(x + y) = f(x) * f(y).[/MATH]
[MATH]a > 0 \text { and } f(x) = a^x \implies f(x + y) = a^{(x+y)} = a^x * a^y = f(x) * f(y).[/MATH]
[MATH]\therefore f(1) = 3 \implies a^1 = 3 \implies a = 3 \implies f(x) = 3^x.[/MATH]
Please state problems exactly.

 

[HallsofIvy]

This qu

We have a function such that [MATH]f(x+y)=f(x*y)[/MATH][MATH]f(1)=3[/MATH]What's [MATH]f(2)[/MATH]?

It seems like it's a function [MATH]3^x[/MATH] but I'm not sure. Then the answer would be 2. Is it correct?

Well, if [math]f(x)= 3^x[/math] then [math]f(2)= 3^2= 9[/math], not 2.

But the problem does not ask for you to find the actual function, f(x), nor does it imply that there is only one. Given that f(x+ y)= f(x*y) then f(2)= f(1+ 1)= f(1*1)=f(1)= 3.

 

[Steven G]

We have a function such that [MATH]f(x+y)=f(x*y)[/MATH][MATH]f(1)=3[/MATH]What's [MATH]f(2)[/MATH]?

It seems like it's a function [MATH]3^x[/MATH] but I'm not sure. Then the answer would be 2. Is it correct?

This could not be any easier. Let x=y=1. Now f(2) = f(1+1)=f(1*1)=f(1) = 3
Oh, HallsofIvy beat me to the punch line.

 

[Darya]

Well, if [math]f(x)= 3^x[/math] then [math]f(2)= 3^2= 9[/math], not 2.

But the problem does not ask for you to find the actual function, f(x), nor does it imply that there is only one. Given that f(x+ y)= f(x*y) then f(2)= f(1+ 1)= f(1*1)=f(1)= 3.

Omgg, thanks! A lil typo with that 2