3f(x)=f(3x+1)-1 -> 3f(2)=f(6+1)-1 -> 3*5=f(7)-1 -> 16=f(7)yellowho wrote:The function f(x) is such that f(3x+1) = 3f(x) + 1 for all values of x. If f(2)=5, then f(7)=?
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Night reader
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Last edited by Night reader on Fri Jan 21, 2011 3:35 am, edited 1 time in total.
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yellowho
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I don't know what you did there but I think you got the relationship backward.
My real question in this is, is there a way to solve this if 3x+1=7 is not true (X is not 2) as defined? Can you look at this like a sequence?
My real question in this is, is there a way to solve this if 3x+1=7 is not true (X is not 2) as defined? Can you look at this like a sequence?
Last edited by yellowho on Thu Jan 20, 2011 11:38 pm, edited 1 time in total.
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Night reader
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f(3x+1) = 3f(x) + 1 not 3x+1=7yellowho wrote:I don't know what you did there but I think you got the relationship backward.
My real question in this is, is there a way to solve this if 3x+1=7 is not true (X is not 2) as defined? Can you look at this like a sequence?
f(2)=5 is different function
we need to find yet another function f(7)
I tried my best, though it's nonstandard algebra application rather than backward restatement.
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Rahul@gurome
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We know f(2) and our target is to determine the value of f(7). How we can do that?yellowho wrote:The function f(x) is such that f(3x+1) = 3f(x) + 1 for all values of x. If f(2)=5, then f(7)=?
One quiet obvious way is to express f(7) in terms of f(2). Can we do that? Yes! From the relations given, we can do that easily. How?
Just put x = 2 in the given relation.
Now we have, f(3x + 1) = 3f(x) + 1
=> f(3*2 + 1) = 3f(2) + 1
=> f(7) = 3f(2) + 1
Now we have f(7) in terms of f(2).
Just put the value of f(2), and we have f(7) = 3*5 + 1 = 16
How will you justify that step?Night Reader wrote:f(2)=f(6+1)=3f(2)+1 = 5 -> f(7)=3f(2)+1-5 -> f(7)=3f(2)-4
What you've wrote is equivalent to f(2) = f(7)
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yellowho
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sorry didnt mean to offend. i just didn't understand what u did. that doesnt mean you are wrong tho. i posted this question to look for different ways to solve so a different approach is welcome.
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Night reader
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thanks Rahul, I have edited my solution.Rahul@gurome wrote:We know f(2) and our target is to determine the value of f(7). How we can do that?yellowho wrote:The function f(x) is such that f(3x+1) = 3f(x) + 1 for all values of x. If f(2)=5, then f(7)=?
One quiet obvious way is to express f(7) in terms of f(2). Can we do that? Yes! From the relations given, we can do that easily. How?
Just put x = 2 in the given relation.
Now we have, f(3x + 1) = 3f(x) + 1
=> f(3*2 + 1) = 3f(2) + 1
=> f(7) = 3f(2) + 1
Now we have f(7) in terms of f(2).
Just put the value of f(2), and we have f(7) = 3*5 + 1 = 16
How will you justify that step?Night Reader wrote:f(2)=f(6+1)=3f(2)+1 = 5 -> f(7)=3f(2)+1-5 -> f(7)=3f(2)-4
What you've wrote is equivalent to f(2) = f(7)
