See Question and working out below. The mark scheme says to read the y intercept as -0.45 however in my working I use lg(0.45) which would surely be correct? (Correct answer on mark scheme is 0.51, I got 0.78) https://gofile.io/d/ZazDLf https://gofile.io/d/Iryn6z
If the value for y-intercept is -0,45 - that is what I got - and your logarithm function y = 0,5*lg(gk/4pi²), the equation has to be:
0,45 = 0,5*lg(gk/4pi²)
Guess that g is a constant.
Thanks very much for your reply, what I don't understand is that if the y axis was in say (x10^3), you would write -0.45x10^-3 as the y-intercept, so how come i don't write lg(-0.45) instead of -0.45?
Thanks very much for your reply, what I don't understand is that if the y axis was in say (x10^3), you would write -0.45x10^-3 as the y-intercept, so how come i don't write lg(-0.45) instead of -0.45?
Do you know what is called linearising an equation? If not, watch the following video. https://youtu.be/mxpvkXvqveQ?si=Ysx85mZVHrmiwFJg (There is a segment of the video (10:01 to 11:11) that you want to skip)
If yes, then you may have forgotten the fundamentals. For example, if we want to linearise the equation y=A∗bx where A and b are constant. The linear form would be
lg(y) = x lg(b) + lg(A)
which looks like Y = mX + c where m is the gradient and c is the y-intercept.
As c is the Y-intercept, so when we read a value say 2.3 on the y-axis when X = 0, c=lg(A)=2.3
One thing to note, you cannot lg a negative value.