Applying a Horizontal Stretch to ln(x): Understanding the Shift in f(x)

  • Thread starter Thread starter andrewkg
  • Start date Start date
  • Tags Tags
    Transformations
AI Thread Summary
Applying a horizontal stretch to f(x) = ln(x) by a factor of k (where k > 1) is equivalent to a horizontal compression by a factor of 1/k, not a shift. To achieve a horizontal stretch, the function should be expressed as g(x) = f(x/k). The discussion highlights the importance of understanding the distinction between horizontal stretches and shifts in logarithmic functions. Additionally, the relationship between logarithmic properties and transformations is emphasized. Overall, clarity on these concepts is crucial for accurate function manipulation.
andrewkg
Messages
86
Reaction score
0
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?
 
Physics news on Phys.org
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
f\left( \frac{x}{k} \right).
 
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.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top