Understanding graphing

Hello there

I wonder if anyone may be able to help me understand how to answer these questions regarding graphing?

Physics 3 1.jpg

For the first graph, I did:

y(0) = 10^0 = 1, y(1) = sqrt(10) = about 3.16, y(2) = 10^1 = 10, y(3) = 10*sqrt(10) = about 31.6, y(4) = 10^2 = 100.

And plotted it as:

Physics 3 1 graph1.jpg

Does this look incorrect?

I'm really unsure what z = log(y) means and therefore what I must do for the second graph.

Please might anyone be able to help me? I really hope to understand.

Thank you very much

Reply 1

davros

Original post by Ggdf

Hello there

I wonder if anyone may be able to help me understand how to answer these questions regarding graphing?

Physics 3 1.jpg

For the first graph, I did:

y(0) = 10^0 = 1, y(1) = sqrt(10) = about 3.16, y(2) = 10^1 = 10, y(3) = 10*sqrt(10) = about 31.6, y(4) = 10^2 = 100.

And plotted it as:

Physics 3 1 graph1.jpg

First graph looks fine. (All graphs of a^x for some a > 0 basically have the same shape)

What work have you done with logarithms? You will have a button on your calculator that calculates logarithms for you, so you should have no problem working out a value for z for each value of x, and therefore plotting z against x.

The later parts of the question are leading you to a result that you should be able to work out if you've seen the laws of logarithms. Do you know what log(a^b) is in terms of log a?

Reply 2

Ggdf

Original post by davros

Thank you very much for your reply.

I understand very little about logarithms, only that the log button on the calculator relates to the base 10 - the number of times 10 must be raised to get the number being logged. I also know that logarithms linearize the plots (although I don't know why) and that using log values on graphs is advantageous as it brings values that would be very far out of range into a manageable scale for analysis. It can also help produce usable trendlines (again I don't know why). I've never really learnt anything about logarithms and would love to understand.

For this exercise, should I log each of the plotted values, i.e. log 1, log 3.16, log 10, log 31.6, log 100?

I really appreciate your help,

Reply 3

davros

Original post by Ggdf

Are you doing this as part of a course? Do you not have a teacher explaining how logs work etc?

Logarithms are always expressed relative to a particular base. You have described the base-10 logarithm which is one of the most common ones.

Taking a logarithm reverses exponentiation, so if

10^a = b

then

log_{10} b = a

Yes, you should plot the logs of the values you calculated earlier.

Check in your book (or on the web) for the laws/rules of logarithms. You should find laws like

log(uv) = log u + log v

and

log(x^r) = rlogx

which should help you answer later parts :smile:

Reply 4

Ggdf

Original post by davros

10^a = b

then

log_{10} b = a

Yes, you should plot the logs of the values you calculated earlier.

Check in your book (or on the web) for the laws/rules of logarithms. You should find laws like

log(uv) = log u + log v

and

log(x^r) = rlogx

which should help you answer later parts :smile:

Thank you so much for your help :smile:

.

Sadly I don't have a teacher. I'm doing a science course which involves a small amount of maths that is done by distance learning. For this I don't have a teacher to consult. I was accepted onto the science course based purely on my science background and abilities, although I have a very weak maths background (due to gaps in my education in the past as a result of health problems). I haven't done anything on logarithms. This maths is very much beyond me, although my course assumes I know it already which is why I don't receive tuition.

I've taken a look at the laws of logarithms, but am still quite confused because I'm also not too familiar with some of the terminology that is used to explain them. It troubles me that I don't understand; it is definitely a topic I would like to learn in detail.

In regard to my question; I've plotted the log10 (on the calculator) values of the y values to get the following graph:

Physics 3 1 graph2.jpg

I understand and see what you describe about the logarithm reversing the exponentiation. I can recognise that the x values were initially multiplied by 0.5 and then each of these values were used to raise 10 to give values to be plotted. I see that taking the logarithm reverses this. I can see that the x axis values initially go up in 1s and that the exponentiation goes up in 0.5s. I can recognise the regularity of this and can vaguely see the logic of why taking the log10 gives a straight line. However my understanding of why we get this straight line isn't clear. I wonder what has happened and how the straight line can be described?

I've calculated the gradient of the line to get 0.5. I've also done dz/dx for each value to get 0.5 for each. I can see that the gradient and dz/dx are the same. I know the gradient shows how much y goes up for each increase in x. I can overall see that that this is x0.5. However, again, I really can't put these ideas together clearly to explain why the line is straight and also how the gradient relates to dz/dx to answer the last 2 questions. Do you think I've in any way described it adequately?

I really am grateful for your help.

Reply 5

davros

Original post by Ggdf

Thank you so much for your help :smile:

.

In regard to my question; I've plotted the log10 (on the calculator) values of the y values to get the following graph:

I understand and see what you describe about the logarithm reversing the exponentiation. I can recognise that the x values were initially multiplied by 0.5 and then each of these values were used to raise 10 to give values to be plotted. I see that taking the logarithm reverses this. I can see that the x axis values initially go up in 1s and that the exponentiation goes up in 0.5s. I can recognise the regularity of this and can vaguely see the logic of why taking the log10 gives a straight line. However my understanding of why we get this straight line isn't clear. I wonder what has happened and how the straight line can be described?

I've calculated the gradient of the line to get 0.5. I've also done dz/dx for each value to get 0.5 for each. I can see that the gradient and dz/dx are the same. I know the gradient shows how much y goes up for each increase in x. I can overall see that that this is x0.5. However, again, I really can't put these ideas together clearly to explain why the line is straight and also how the gradient relates to dz/dx to answer the last 2 questions. Do you think I've in any way described it adequately?

I really am grateful for your help.

You've pretty much solved this for yourself :smile:

As I said earlier,

10^a = b \Rightarrow log b = a

if we're taking base-10 logarithms.

So if

y = 10^{0.5x}

then taking logarithms tells us that

log y = 0.5x

. But since we defined z = log y, we can write z = 0.5x, which is just the equation of a straight line with gradient 0.5 passing through the origin.

Try and find some more material on logarithms and exponentials to read through as these topics underpin a lot of scientific laws,

Reply 6

Ggdf

Original post by davros

You've pretty much solved this for yourself :smile:

As I said earlier,

10^a = b \Rightarrow log b = a

if we're taking base-10 logarithms.

So if

y = 10^{0.5x}

then taking logarithms tells us that

log y = 0.5x

I can't tell you how much I appreciate your guidance with this :smile:

.

When relating the gradient to dz/dx, should I say that dy/dx = 0.5 in all cases and means that each x was multiplied by 0.5. Therefore, like the gradient, it shows that y goes up by 0.5 with each increase in x?

Sorry if I'm mistaken.

Reply 7

davros

Original post by Ggdf

I can't tell you how much I appreciate your guidance with this :smile:

Be careful!

Plotting z against x gives a straight line, and dz/dx = 0.5.

However, y plotted against x isn't a straight line and it's not true that dy/dx = 0.5!

Reply 8

Ggdf

Original post by davros

Be careful!

Plotting z against x gives a straight line, and dz/dx = 0.5.

However, y plotted against x isn't a straight line and it's not true that dy/dx = 0.5!

Oh, I'm very sorry, that was my mistake (a typo)! I see why that's wrong - thank you for pointing it out! But the general idea is OK?

When relating the gradient to dz/dx, should I say that dz/dx = 0.5 in all cases and means that each x was multiplied by 0.5. Therefore, like the gradient, it shows that y goes up by 0.5 with each increase in x?

Is the y highlighted (in bold) incorrect?

Reply 9

davros

Original post by Ggdf

But it's not "y" that goes up by 0.5 with every x, it's "z"!

Reply 10

Ggdf

Original post by davros

But it's not "y" that goes up by 0.5 with every x, it's "z"!

Yes, I see! Thank you so much, you've been fantastic! :smile:

Reply 11

Ggdf

Original post by davros

You've pretty much solved this for yourself :smile:

As I said earlier,

10^a = b \Rightarrow log b = a

if we're taking base-10 logarithms.

So if

y = 10^{0.5x}

then taking logarithms tells us that

log y = 0.5x

Hi Davros

I'm sorry to trouble you. I wonder, should I label the vertical and horizontal axis of this graph at all? I'm unsure what the correct labels would be? do you think x would be 'x' and y be 'y=0.5x'?

Thank you very much.

Reply 12

davros

Original post by Ggdf

Well it depends what you're plotting :smile:

If you plot y against x, label the axes x and y; if you're plotting z against x. label the, x and z. It is also usual to write the equation of the graph somewhere across it near the line/curve e.g. y = 10^x or z = 0.5x or whatever.

Reply 13

atsruser

Original post by Ggdf

I'm really unsure what z = log(y)

There is a point that some people never seem to grasp: the word "logarithm" merely means "power".

So if we write

4^2

we call 4 the "base" and 2 the "power" or the "logarithm". We know that

4^2 = 16

so we could say "the power of 16 using the base 4 is 2". We don't, however; we say "the logarithm of 16 to the base 4 is 2".

We can ask the question: what power do I need to turn 4 into 16? Mathematicians have created a function called

\log

that answers this question. We feed a number into a

\log

function and it spits out the power. There is a *different*

\log

function for each base and each

\log

function only knows about its own base e.g.

\log_4(16) = 2

because

4^2=16

\log_3(27) = 3

because

3^3=27

Generally we write the function like so

\log_4(x)

for some arbitrary number

x

whose power of 4 we want.

We can be nice to a

\log

function by feeding in a number already expressed as a power of the base:

\log_4(4^2) = 2

Now it's obvious what number the function should spit out.

Note that there is a 2 inside the brackets and one outside the brackets. We can understand that by thinking about another function, namely the "4 to some power" function. This function eats a number (the power or the logarithm - they're the same thing, remember) and spits out the result of raising the base to that power. We write this function as

4^x

and we used it earlier where we saw that

4^2=16

.

Now we can see that if we start with 2, then feed that into the

4^x

function, then feed the result of *that* into the

\log_4

function, we get 2 back again. That means that the

\log_4

function undoes the work of the

4^x

function i.e. they are inverses. In general:

\log_4(4^x)=x

That works the other way too. Using the

\log_{10}

function we have:

10^{\log_{10}(1000)} = 1000

since

\log_{10}(1000)=\log_{10}(10^3) = 3

10^{\log_{10}(1000)} = 10^3= 1000

.

If we start with the

\log_4

function and try the same trick, we can discover something interesting:

4^{\log_4(2)} = 2

This is true since the functions are inverses. But note that we have something of the form

4^x=2

. If you about powers, you know that means that

x=\frac{1}{2}

. So we now know that:

\log_4(2)=\frac{1}{2}

(We could figure this out in other ways too though). But note that

\frac{1}{2} = \frac{1}{\log_2(4)}

\log_4(2)=\frac{1}{\log_2(4)}

. A little thought shows that this is a general rule:

\log_a(b) = \frac{1}{\log_b(a)}

(Note also that we can now draw the graph of a

\log

function. Since it is the inverse of a power function, we merely reflect the power function curve in the line

y=x

to get the corresponding

\log

graph.)

Something very nice and useful happens with

\log

functions and products. Note that:

a^p \times a^q = a^{p+q}

and that

\log_a(a^p) = p, \log_a(a^q)=q

and that

\log_a(a^{p+q})=p+q

. We can write that last equation another way:

\log_a(a^p\times a^q) = \log_a(a^p) + \log_a(a^q)

and we can simplify that a bit by saying

c=a^p, d=a^q

so that:

\log_a(cd) = \log_a(c)+\log_a(d)

That says that the

\log

function turns a product into a sum. This is the basis for the operation of a slide rule. The slide rule had numbers positioned where their logs would be. If you then added the position of, say, 3.56, to the position of, say, 25.78, you were adding their logs i.e. you were multiplying the numbers 3.56 and 25.78, and you could read off the result of this product on the rule without doing a difficult calculation. Casio killed all that niceness. Bastards.

Anyway, that's a brief summary of logs and how they work. There's more to it, but you can figure it out using similar reasoning. Remember: it all follows from this equivalence: logarithm means power.