Math question

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Farcus

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F(x)=3^x and F(a)=5 what is F(2a)

ok i saw the solution and the first part makes sense where it is 3^2a but what property is it that allows you to make it (3a)^2? I don't remember doing this. Also then the solution says (3a)^2 = [f(a)]^2 what is this? how is F(a) = 3a now? wtc?!
 
F(x)=3^x and F(a)=5 what is F(2a)

ok i saw the solution and the first part makes sense where it is 3^2a but what property is it that allows you to make it (3a)^2? I don't remember doing this. Also then the solution says (3a)^2 = [f(a)]^2 what is this? how is F(a) = 3a now? wtc?!
Is the answer = 25? If yes, then I'll show you my way of solving it. No log necessary.
 
F(x)=3^x and F(a)=5 what is F(2a)

ok i saw the solution and the first part makes sense where it is 3^2a but what property is it that allows you to make it (3a)^2? I don't remember doing this. Also then the solution says (3a)^2 = [f(a)]^2 what is this? how is F(a) = 3a now? wtc?!

...okay so I didn't work this out all the way so I'm not totally sure it will give you the right answer, but I am almost positive you use log. This is what I was kind of playing with...
Log(base a)B=C is the same as B=a^(C) so...
if F(x)=3^(x) and F(a)=5, then F(a)=3^(a) so....
3^(a)=5 then Log(base 3)5=a so... find out what a equals and you get....
3^(2a)=5 then LOG(base 3)?=2a

You can also write F(2a) out as 3^(2a) which becomes 3^(2)*3^(a) by properties of exponents.

I'm sure there's some property that would help you better but I can't think of it. I hope I didn't confuse you more, just offering some different ways to manipulate the problem. Don't know if it helps any but it's worth a shot...
 
F(x)=3^x
F(2a)=3^(2a)
3^(2a)=(3^a)^2
If 3^a=5, then 3^(2a)=5^2=25

Yes, 25 is right. Logs may work as well, but this was easier for me. I'm sure you did it this way as well.
 

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