# Logarithm Graph Question

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
(edited 3 months ago)
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.
(edited 3 months ago)
Original post by Kallisto
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?
You do but it is (gk/4pi²)^0.5 = -0.45
You can't equate logs without bringing the power up first
Original post by crispychips24
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 $y = A*b^x$ 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.