# Transformations math help

1. Mar 14, 2013

### andrewkg

Q
Applying a horizontal stretch by a factor of k (where k is a constant such that k>1) to f(x)=lnx is equivalent to applying what shift to f? Give both the amount and direction of the shift.

my A
so i came to the conclusion that the answers must have to do with the laws of logs. and from that i cam to the conclusion the shift = to f(kx)=ln(kx)=ln(x)+ln(k) are = so the shift of f(x) would be f(x)+ln(k).

What do you guys think?

2. Mar 14, 2013

### Fredrik

Staff Emeritus
That looks good. Of course, my understanding of what you mean by "stretch" and "shift" is based on the answer you came up with, so it's not like I could look at the problem, solve it and then compare my result to yours.

3. Mar 14, 2013

### andrewkg

Re: Transformations

Well vertical shift ment up or down the desired unit. And horazontal stretches and compressions. By the desired factor. Hmm not sure how to put that. Well that basically what the book says.

4. Mar 14, 2013

### eumyang

Actually, if k > 1, then f(kx) is a horizontal shrink of f(x) by a factor of 1/k. If you want a horizontal stretch by a factor of k, with k > 1, then you should write it as
$f\left( \frac{x}{k} \right)$.

5. Mar 14, 2013

### Fredrik

Staff Emeritus
Your answer to the problem gave me more information than that. The horizontal stretch by a factor k is presumably the map $f\mapsto g$ where g is defined by g(x)=f(x/k) for all x.

6. Mar 14, 2013

### andrewkg

thanks you guys. Once again PF has saved me from a careless error.