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What's the quickest method of determining where on this scale is, for example, 1.5 or 5, etc?
Brainiacs needed!
- Thread starter moribundman
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It's an exponential scale of 2. 1,2,4,8,16,32,64,128 etc...
Divide by Pi
Give a wrong answer then wait for you to correct it.
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Extrapolate half the distance between 4 and 8, which would be 6 and then extrapolate between 4 and 6 to locate 5. Same thing to find 1.5 which is halfway between 1 and 2. That is how I would do it.
Originally Posted By: SaturnIonVue
Extrapolate half the distance between 4 and 8, which would be 6 and then extrapolate between 4 and 6 to locate 5. Same thing to find 1.5 which is halfway between 1 and 2. That is how I would do it.
+1. I like that answer. I'd do the same.
But, I'll be the first to admit I'm no "brainiac".
Extrapolate half the distance between 4 and 8, which would be 6 and then extrapolate between 4 and 6 to locate 5. Same thing to find 1.5 which is halfway between 1 and 2. That is how I would do it.
+1. I like that answer. I'd do the same.
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I just don't like the sloppy way you wrote the numbers. The line is ....acceptable
(I, however, AM left handed
- the non-Spaniard swordsman who was observing off camera in Princes Bride)
btw- are you really asking the quickest way to determine it ..or the most intelligent and succinct way to express it?
(I, however, AM left handed
btw- are you really asking the quickest way to determine it ..or the most intelligent and succinct way to express it?
Quote:
Hmmmm log table base 2
Shannows got it. An exponential scale becomes is linear when you use the logarithm.
The log scale will read 0 1 2 3
(2^0=1, 2^1=2, 2^2=4, 2^3=8)
You then use the chart to determine 2 to the ? power = 1.5 or 5
The extrapolation method (really interpolation) doesn't work since the orginal scale is not linear.
Hmmmm log table base 2
Shannows got it. An exponential scale becomes is linear when you use the logarithm.
The log scale will read 0 1 2 3
(2^0=1, 2^1=2, 2^2=4, 2^3=8)
You then use the chart to determine 2 to the ? power = 1.5 or 5
The extrapolation method (really interpolation) doesn't work since the orginal scale is not linear.
Last edited:
JHZR2
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Originally Posted By: Shannow
Hmmmm log table base 2
+1
Don't necessarily see the need for much brain on this
Hmmmm log table base 2
+1
Don't necessarily see the need for much brain on this
The easiest way...
Google it
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Originally Posted By: simple_gifts
Quote:
Hmmmm log table base 2
Shannows got it. An exponential scale becomes is linear when you use the logarithm.
I cannot use a linear scale. It has to be an exponential scale.
Quote:
Hmmmm log table base 2
Shannows got it. An exponential scale becomes is linear when you use the logarithm.
I cannot use a linear scale. It has to be an exponential scale.
You do it with calculus...
(I have no idea, but stuff like this always comes down to somebody pulling out a crazy calculus explanation.)
(I have no idea, but stuff like this always comes down to somebody pulling out a crazy calculus explanation.)
You need a Japanese person to help you, they are real smart with Calculus.
Google that $#!@
Google that $#!@
Quickest, exponential scale...are you going to make another Hubble telescope out of a pair of binoculars?
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Massage what I'm trying to say for the sensible platform.
Divide it into 15 parts. 7.9-8.0 will be 1/15 of the total distance.
This may be off since we didn't start @ ZERO.
I'm envisioning more of an amortization chart. Between each integer isn't evenly spaced for value (for lack of a better word.
1.5 does not reside equidistant between 1 and 2.
Divide it into 15 parts. 7.9-8.0 will be 1/15 of the total distance.
This may be off since we didn't start @ ZERO.
I'm envisioning more of an amortization chart. Between each integer isn't evenly spaced for value (for lack of a better word.
1.5 does not reside equidistant between 1 and 2.
Originally Posted By: moribundman
Originally Posted By: simple_gifts
Quote:
Hmmmm log table base 2
Shannows got it. An exponential scale becomes is linear when you use the logarithm.
I cannot use a linear scale. It has to be an exponential scale.
I thought this was a riddle instead of a serious question. My bad.
n=logx/log2 (this avoids messing around with base2 although a base2 solution would be more elegant...if you consider something from old High School trig elegant.
Where n = number of ticks over on your scale with the first tick being the 0th tick.
and x = the number you want to add to your scale.
Lets try 4. n=logx/log2=log4/log2 =2 which is the second tick over on your scale and it is 4.
Now try 5. log5/log2=2.3219, so 5 is 2.3219 ticks over on your scale.
Originally Posted By: simple_gifts
Quote:
Hmmmm log table base 2
Shannows got it. An exponential scale becomes is linear when you use the logarithm.
I cannot use a linear scale. It has to be an exponential scale.
I thought this was a riddle instead of a serious question. My bad.
n=logx/log2 (this avoids messing around with base2 although a base2 solution would be more elegant...if you consider something from old High School trig elegant.
Where n = number of ticks over on your scale with the first tick being the 0th tick.
and x = the number you want to add to your scale.
Lets try 4. n=logx/log2=log4/log2 =2 which is the second tick over on your scale and it is 4.
Now try 5. log5/log2=2.3219, so 5 is 2.3219 ticks over on your scale.
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Originally Posted By: XS650
n=logx/log2
That's what I couldn't remember or figure out on my own! Yikes. Thanks for jogging my memory!
n=logx/log2
That's what I couldn't remember or figure out on my own! Yikes. Thanks for jogging my memory!
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