noob phys question - Page 2

Purind

Canada3562 Posts

September 17 2008 04:53 GMT

#21

On September 17 2008 12:03 nAi.PrOtOsS wrote:
Ok so my teacher says that as long as the log-log graph is linear then it's correct and you can find the slope and then use that in other calculations. Is it possible tho, to have a linear log-log graph but be completely wrong? so the slope of that incorrect linear line would completely mess up all the calculations?

Depends on what you're graphing.

If it's supposed to be exponential, then I think it should be linear on a log log

Underwhelmed

United States207 Posts

September 17 2008 06:30 GMT

#22

I assume you have taken data points and are plotting them on a log(x) vs log(y) graph. So basically, log(y_2) - log(y_1) / log(x_2) - log(x_1).

Ilikestarcraft

Korea (South)17719 Posts

September 17 2008 07:04 GMT

#23

On September 17 2008 12:53 micronesia wrote:
What percent of the people here responded while having no idea what log-log means?

+ Show Spoiler +

OVER 9000!

i dont know what log log is im just sortof annoyed because ops been making a lot of threads like this lately. Im not saying asking for help is bad i used to make threads like these too.

jgad

Canada899 Posts

September 17 2008 07:39 GMT

#24

On September 17 2008 13:20 IzzyCraft wrote:

Show nested quote +

Doesn't matter slope is always best fit line or learn how to write algorithms.

That's not true. If you plot a straight line (linear function) on a log-log graph you get a straight line with a slope of 1, regardless of what the slope would be on a linear-linear graph. If you plot a quadratic equation on a log-log graph, you get a straight line with a slope of 2, not a curving line. My guess is that this exercise is about learning to interpret the shape of plots in logarithmic form and to understand the congruency between different forms of the same plot.

jgad

Canada899 Posts

September 17 2008 07:43 GMT

#25

On September 17 2008 13:53 Purind wrote:

Show nested quote +

Depends on what you're graphing.

If it's supposed to be exponential, then I think it should be linear on a log log

Nah, exponential still curves on a log-log graph. It's a straight line on a log-lin graph.

sigma_x

Australia285 Posts

September 17 2008 07:43 GMT

#26

to find the gradient of a log log graph, put your points on log log graph paper, and find the slope. Since this is physics, (and points won't lie neatly on a linear graph) doing it any other way will not get you the answer, unless finding L1 approximations using the [dual of a] simplex is your cup of tea.

"Ok so my teacher says that as long as the log-log graph is linear then it's correct and you can find the slope and then use that in other calculations. Is it possible tho, to have a linear log-log graph but be completely wrong? so the slope of that incorrect linear line would completely mess up all the calculations?"
This will depend on what your actually trying to find. This will be fine if you the actual question features a linear component in logs. For example, y=1/2 g t^2 [which is the distance travelled by a falling object due to gravity] can be put in the form y=mx+b by taking logs of both side.

so, logy=2(logt) +log(1/2 g), which is linear on a log log graph.

Notice that in this example, the gradient tells you the power of "t", so if your gradient is roughly "2" it shouldn't interfere with your calculations too much.

Maenander

Germany4919 Posts

September 17 2008 07:56 GMT

#27

On September 17 2008 12:53 micronesia wrote:
What percent of the people here responded while having no idea what log-log means?

+ Show Spoiler +

OVER 9000!

If the OP asks stupid questions, he gets stupid answers. That´s perfectly fine.

fight_or_flight

United States3988 Posts

September 17 2008 08:54 GMT

#28

Would it be something like

log(y) = slope*log(x) + intercept
10^log(y) = 10^(slope*log(x) + intercept)
y = x^slope*10^intercept

y = (10^intercept)*x^slope

?

micronesia

United States24502 Posts

September 17 2008 10:19 GMT

#29

On September 17 2008 16:56 Maenander wrote:

Show nested quote +

If the OP asks stupid questions, he gets stupid answers. That´s perfectly fine.

He's asked too many questions lately, and the OP was too short, and he could have been more clear, but it wasn't actually stupid imo. Not his fault people don't know what log-log is but reply anyway.

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