Posted by marcus on Wednesday, November 21, 2012 at 3:18am.
g-1 = -e^(x+1)
1-g = e^(x+1)
x + 1 = ln (1 - g)
x = ln(1-g) -1
g will always be less than 1.
Is that what you want? I don't know what you mean by "exponential form"
You can always sketch the graph using the original equation. Just assume different x values, compute g for each, and plot the result
My book says a way for me to graph/plot points for my function is to write it in exponential form. For example. "Graph (fx)= log base 2(x-2)
y=log base 2 (x-2)
x-2=2^y
x=2^y + 2
So now I can just plug any y value in to get my x value to graph my translated graph. So I was wondering on how to be able to put the above function in exponential form as well? I know when you're done with the graph it's supposed to go through the point (-1,0) and the asymptote is y=1, but I don't know how to get there.
Related Questions
precalculus - solve exponential equation. express in exact form. e^x-42e^-x= -19
Algebra Help - Let x = log_2 1/8 Write the exponential form of the equation and ...
precalculus - solve the exponential equation express in exact form only show all...
Algebra II - 1) There is a solution to x = -(y-5)^2 and x = (y -5 )^2 + 7 a) ...
Math - How do we turn this into exponential form? cubert[125 q^9 s^15] Any tips ...
Precalculus - "Sketch the ellipse. Find the coordinates of its vertices and...
Algebra 1 - A line passes through (4,2) and is parallel to the graph of x= -3. ...
precalculus - Find the focus, directrix, and focal diameter of the parabola. x2...
Calculator reposting - Hi, I'm trying to graph the equation "y=3x"...
Calculator problem-- urgent!!! - Hi, I'm trying to graph the equation "...
For Further Reading
