hi we have been given a graph and we need to fing its equation. the graph is logistic. we are asked to take any six points on it and then apply appropriate log to get a straight line. i don't know how to get that straight line?
i just know for exponential logy vs x gives a straight line.
To convert a logistic curve into a straight line, you can take the logarithm of both sides of the logistic function. The most common logarithm used for this purpose is the natural logarithm, denoted as ln.
The logistic equation is typically expressed as:
y = K / (1 + Ae^(-Bx))
To convert this equation into a straight line, you can take the natural logarithm of both sides:
ln(y) = ln(K / (1 + Ae^(-Bx)))
Using logarithm properties, we can simplify this equation:
ln(y) = ln(K) - ln(1 + Ae^(-Bx))
Now, let's introduce a new variable, say "z," which is equal to ln(y). We can rewrite the equation as:
z = ln(K) - ln(1 + Ae^(-Bx))
Now, this equation is in the form of a linear equation (y = mx + c), where z is equivalent to y, ln(K) is the y-intercept (c), -ln(A) is the slope (m), and Bx is x.
To get a straight line from this equation, you can plot the points (x, z) using six chosen pairs of (x, y) values from the original logistic curve. Then, use these points to draw a straight line on a graph.
Remember that in order to find the equation of the straight line from the graph, you need to determine the slope (m) and the y-intercept (c) by calculating the corresponding values from the coordinates of any two points on the line.
Once you have determined the values of m and c, the equation of the straight line in terms of z (ln(y)) will be:
z = mx + c
Finally, you can replace z with ln(y) to get the equation in terms of y:
ln(y) = mx + c
Remember to use the base of the logarithm you initially chose (in this case, natural logarithm, ln) to convert back to y if necessary.