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

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.

-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.

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

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.

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