Horizontal stretch of ln(x) by factor k equivalent to what shift

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andrewkg
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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?
 
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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.
 


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.
 
andrewkg said:
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?

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
[itex]f\left( \frac{x}{k} \right)[/itex].
 
andrewkg said:
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.
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.
 
thanks you guys. Once again PF has saved me from a careless error.