BMAD Posted September 22, 2017 Report Share Posted September 22, 2017 Find a continuous function where the following identity is true: f(2x) = 3f(x) Quote Link to comment Share on other sites More sharing options...

1 Buddyboy3000 Posted September 24, 2017 Report Share Posted September 24, 2017 Spoiler I'm not too sure if this is right, but I got two possible solutions: y=3^(log(x)/log(2)) and y=-3^(log(-x)/log(2)) I got these two solutions by trying to make an equation where the y-value triples only when the x-value doubles, and not after each x-value. Quote Link to comment Share on other sites More sharing options...

1 ThunderCloud Posted December 5, 2017 Report Share Posted December 5, 2017 You asked for it... Spoiler f(x) = 0 Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted December 7, 2017 Report Share Posted December 7, 2017 Sort of combining the two solutions already given Spoiler f(x) = C 3^{log}^{2}^{x} where C is any constant f(2x) = C 3^{log}^{2}^{(2x)} = C 3^{1+log}^{2}^{x} = C (3 3^{log}^{2}^{x}) = 3f(x) Quote Link to comment Share on other sites More sharing options...

2 ThunderCloud Posted December 7, 2017 Report Share Posted December 7, 2017 (edited) 25 minutes ago, plasmid said: Sort of combining the two solutions already given Hide contents f(x) = C 3^{log}^{2}^{x} where C is any constant f(2x) = C 3^{log}^{2}^{(2x)} = C 3^{1+log}^{2}^{x} = C (3 3^{log}^{2}^{x}) = 3f(x) Spoiler I suspect the issue with this solution is that f(x) is not defined for all x, and therefore not continuous for all x. f(x) = 3^{log2(|x|)}almost works, but still has a discontinuity at x = 0. Edited December 7, 2017 by ThunderCloud Quote Link to comment Share on other sites More sharing options...

## Question

## BMAD

Find a continuous function where the following identity is true: f(2x) = 3f(x)

## Link to comment

## Share on other sites

## 4 answers to this question

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.