Finding the image point on an inverse Exponential Function.

• Dec 5th 2013, 04:37 PM
sakonpure6
Finding the image point on an inverse Exponential Function.
There is a point (1,0) of the parent function f(x) = log x find the image point on f(x)= 5 log(x-2) which is the same as y= 5f(x) . I know how to do this by applying appropriate transformations, however how come when I use the the algebraic approach found here http://mathhelpforum.com/algebra/203...tml#post737729 , it does not work?

The answer is I found is (0,3).
Thank you.
• Dec 5th 2013, 05:19 PM
HallsofIvy
Re: Finding the image point on an inverse Exponential Function.
No, if you are taking f(x)= log(x), y= 5 log(x- 1) is NOT "y= 5f(x)". It is, rather, y= 5 f(x- 1). Generally speaking changes inside the "parent" function are changes to x and reflect horizontal changes to the graph. For example, f(1)= log(1)= 0 so that f(2)= log(2-1)= 0 so that, in effect, the graph is shifted one place to the right. Changes outside the parent function are changes to y, vertical changes. Multiplying by 5 stretches the graph vertically by a multiplier of 5.
• Dec 5th 2013, 05:57 PM
sakonpure6
Re: Finding the image point on an inverse Exponential Function.
O sorry that was a typo, I know what you mean, but If you actually try solving the problem using the method mentioned in the link it does not work. Am I missing something?