Solve exp(x)=x^4 with a Slide Rule

[TheFerruccio]

I would have replied to the older thread but it seems that is not possible, so I will have to post my question here.

https://www.physicsforums.com/threads/slide-rule.245855/

@BobG mentioned being able to solve a particular problem:

exp(x)=x^4

really quickly using a slide rule. He emphasized how easy it is but I've searched all around the Internet, having used a Versalog slide rule for years, and while I can solve exp(x)=x^3 iteratively using the cube scale, I do not know the quick trick to solving for one power higher.

Maybe it is obvious and I am missing it. Could anyone help me out here?

 

[jedishrfu]

You need to have the loglog scales on your slide rule.

This may help:

http://www.antiquark.com/sliderule/sim/sr-calcs-by-example.html

and more specifically:

http://www.antiquark.com/sliderule/sim/sr-calcs-by-example.html#mozTocId676447

 

[BobG]

Take the natural log of both sides:

e^x=x^4
x = 4 ln x

Now I'll bet you can find the solution(s).

 

[jedishrfu]

He also might be referring to using both indices to find the answer.

Say you want to find the sqrt of 2 then youknow there's an x such that x * x = 2 so on the slide rule C and D scales you slide them back and forth until the number under the 1 index is the same as the number above the 2 that number will be the sqrt of 2. I did that once to compute Ph values but using the loglog scales and the C scale.

 

[BobG]

He also might be referring to using both indices to find the answer.

Say you want to find the sqrt of 2 then youknow there's an x such that x * x = 2 so on the slide rule C and D scales you slide them back and forth until the number under the 1 index is the same as the number above the 2 that number will be the sqrt of 2. I did that once to compute Ph values but using the loglog scales and the C scale.

Exactly. Set up the scales so the answer to any number on the C scale divided by 4 appears on the D scale. And the natural log of any number on the log log scales also appears on the D scale. Somewhere, the number on the C scale is the same as the number on the log log scale.

It does help to sketch the curves for each to figure out how many solutions to expect, but finding the solutions once you know how many you're looking for is easy.

 

[TheFerruccio]

Take the natural log of both sides:

e^x=x^4
x = 4 ln x

Now I'll bet you can find the solution(s).

So that was the first thing I did when posed with this problem but there is probably a mental block that I am missing that is completely obvious here. Even with the explanations I still don't understand what is being done. I understand the case of finding sqrt(x) using just the C and D scales but something that might be obvious is preventing be from taking that mental leap to the case with this problem.[edit]

Oh, dang. It definitely was a mental block. It seems I had the right idea that it was an iterative approach. I ended up getting something like 8.6 and 1.43. By sliding CF to be 4x D, I needed to find the values on CF that matched all the values on the LL scale. So I matched 8.6 on the LL3 scale and 1.43 on the LL2 scale. Lots of flipping over since it's a Versalog (1972, one of the last models). Thanks for the help!