I've forgot to mention arbitrary powers of numbers smaller than one
which is maybe the most obvious use in most instructions.
Let's look on 15*(0.5^3).
If we only have LL1-LL3 then we need to calculate 15/((1/0.5)^3) which
involves at least one value transfer more than using inverse LogLog scales:
0.5 on C to 10 on D; at C1 read 2 on D (Calculate inverse of 0.5)
Cursor to 2 on LL2, 10 on C to cursor; at 3 on C: 8 on LL3
C8 to 10 on D; (at 1 on C 0.125 on D); at 15 on C 1.875 on D
With inverse loglog scales it gets slightly simpler, with only one
instead of two value transfers:
Cursor to 0.5 on LL02; C10 to cursor; at 3 on C: 0.125 on LL03 (going to
left means her to switch to the scale LL0x+1)
C1 to 0.125 on D; at C15: result 1.875 on D
Remark: I strictly use C and D because I use a Studio for these recipes.
If you use a slide rule with inverse and folded scales on its loglog
side you also may use these scales of course. Unfortunately DI scale
virtually never is on LogLog side, so that when starting with an inverse
number is appropriate then the old fashened replacement needs to be used
by setting this number on C to one of the indices of D.
Remark: Intermediate results on LL scales always need to be transferred
for following calculations.