# How to Solve aB=a^B in Neutral Operations Using Algebra

Two Methods:The tutorialExtra Credit Assignment

This article shows how to solve a*b=a^b, which is a Neutral Operation since the operators are being held neutral to one another using the same variable values for a and b. Simple algebra will be used. If a*b represents Area, and a^b represents an elongated graceful exponential Curve, perhaps Mother Nature has a choice / tradeoff / "tipping point" when certain numbers are in this relation. Possibly so. Learn how to determine this relation in a few algebraic steps for your own usage and theoretical understanding.

## Steps

• Become familiar with the basic image of the concept, i.e. when a*B = a^B, then B^(1/(1-B)=a, or for x*Y=x^Y then x = y^(1 (y 1))) ### Method 1 The tutorial

1. 1
Create a new Excel worksheet and enter into cell A1 a*b = a^b, where "^" is Excel's symbol for exponentiation.
• Enter into cell A2 ab/a = (a^b)/a
• Enter into cell A3 b = a^(b-1)
• Enter into cell A4 b^(1/(b-1)) = a; a has now been isolated and defined in terms of b and 1 but b may not = 1 lest division by 0 result in the denominator.
• The simple algebra is complete.
2. 2
Substitute in trial values for b in order to obtain the value of a;
• Let b = 10. Enter in cell C1 ="a=b^(1/(b-1))"
• Enter into cell F1 the word "Var b" for variable b.
• Enter into cell F2 the value 10.
• Do MenuCommand Insert Name DefineName "Var" to cell F2.
• Enter into cell C2 the formula, without quotes, "=Var^(1/(Var-1))"
• Copy the value in cell C2 and do Edit Paste Special Value into cell C3.
• The value you obtain should be 1.29154966501488
3. 3
Test the hypothesis of the original formula, i.e. that a*b = a^b:
• Enter a*b into cell F3
• Enter the formula, without quotes, "=C2*Var" into cell F4.
• Enter a^b into cell F5
• Enter the formula, without quotes, "=C2^Var" into cell F6.
• Enter the formula, without quotes, "=F4-F6" into F7. The result should be 0. Done, because C2*Var = C2^Var, and the difference in F7 is 0!

### Method 2 Extra Credit Assignment

1. 1
It is left to the ingenuity of the reader to prove to themselves, that for Mother Nature, this equation represents a tradeoff or tipping point between area (multiplication) and curve extension (exponentiation) by:
• Copying the above section to an area below it;
• Doing Edit Replace Replace All a with x, b with y, c with z and Var with Var2;
• Defining properly the Var2 cell with the Defined Name Var2 via Insert Name Define;
• Creating a column of the given variable and the dependent formula and filling down the formula, then entering close values for the given variable, probably with an Edit Series menu command;
• A little hard, but do-able: Create the x*y rectangular areas for a chart in terms of lines extending from {0, 0}. For example, if x*y (sub 1) in the first case = 200 and x*y (sub 2) = 202 in the 2nd case, you might create rectangle 1 that's 20*10 and rectangle 2 that's 20.5*(202/20.5=9.85365853658537), or rather, whatever works in the algebraic relation of a*b = a^b, certainly -- one variable should be incremented by an Edit Fill Series menu command is more to the point, and the other left to be computed by the neutral operation;
• And then also charting x^y columnar data as a series on the same chart, by pasting in the data and editing the series to correctly reference the columns and rows desired on the worksheet.
2. 2
Then you will see Nature's "Tipping Point" or Tradeoff, given the equality of resources (numerical amounts of something x and something y).

## Tips

• It is generally a good idea to format INPUT CELLS always as the same color; typically, canary yellow was used in the past. Format Color for cell F2 Canary Yellow. Save the worksheet as "a*b = a^b, Neutral Operation".
• It may be that a Neutral Operation numeric set represents a "tipping point" in Nature. In the above case, she may decide whether to create an area (by multiplying) or an exponential curve (such as an asymptotic stem of a plant, or a growth curve such as e^x).
• n.b. A new Part, The Quadratic Relation of Neutral Operations / Symmetry by Commutation, has been added to the article How to Convert a Quadratic Formula to Roots Form by Completing the Square
• Let E = mc^n as n --> 2. Then E/m = c^n = cn by the Neutral Operation described above. cn = c^n; cn/c = (c^n)/c; n= c^(n-1); n^(1/(n-1)) = c and if c = 186,100, then n = 5.37380552441777E-06 (done by goal seeking in XL). Or if c = 1, then n = 22,751,438,016,029,000 .. per goal seeking. It seems that the operator that produces n=2 is neither exponentiation or multiplication, in truth. "-->" means "approaches". Just for fun, 22,751,438,016,029,000*5.37380552441777E-06 = 122,261,803,299 and 22,751,438,016,029,000^5.37380552441777E-06 = 1.00020241629592 but

186100*5.37380552441777E-06=1.00006520809415=186100^5.37380552441777E-06 but 1*22,751,438,016,029,000 ≠ 1^22,751,438,016,029,000, which I cannot explain, as c*n should equal c^n. Probably for c=1, it simply will not work, so setting c (or c^2) equal to 1 is unjustified according to this. Light speed c has a definite, quantifiable velocity and c^2 a rate of acceleration, if I understand correctly. Perhaps I do not, however.

• For n^(1/(n-1)) = c where n=2, we have 2^(1/1). Where n = 3/2, we have (3/2)^(1/(1/2))= 2.25, or E/m = c^(9/4) and so E/m > c^2, which is interesting. And for n = 5/4, we have (5/4)^(1/(5/4 - 1)) = 625/256 = 2.44140625 -- also interesting.