Simplification steps for manual Newton Raphson iteration on x^(x-a)=b

An easier way to manually adjust the iteration steps used by Newton Raphson you can use the following method for: 
y = xi ^ ( xi - a ) =  b

Use:

xn = xi - [ 1 - b / (xi ^ ( xi - a ))] / [ ln (xi) + 1 - a / xi ]

See sketch. If using a calculator memorize the iterated answer in order to speed up the manual process. 

For instance: 
x^(x-8) = 88 
We guess for the upper solution:
x=9          y = 9^1   = 9 
x=10        y = 10^2 = 100
So x(guess) is for instance x0 = 9.5
Filling in previous gets: x0 = 9.5, a=8, b=88 (using a standard calculator):
x1 = 10.33238273
x2 = 10.09001685
x3 = 9.971697611
x4 = 9.949483974
x5 = 9.948825612 
As the last change is very small, then we check the f(x5) = x5^(x5-8) = 88.000122
This is 88.000122/88-1 = (<0.0‰)

We guess for the lower solution:

x=0.01     y = 9.55 E15
x=1          y = 1

So x(guess) is for instance x0 = 0.5

Filling in previous gets: x0 = 0.5, a=8, b=88

x1 = 0.532744489

x2 = 0.546495332

x3 = 0.548317150

x4 = 0.548344625

As the last change is very small, then we check the f(x4) = x4^(x4-8) = 88.00000826
This is 88.00000826/88-1 = (<0.0‰)

Solution: 

x(lower) = 0.548344625

x(upper) = 9.948825612

In stead of writing long steps, you can fill it in directly. If you implement this in excel, then it might even be way quicker.

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If you have a formula like a^x+bx=c
then you can simplify it by:
x = c/b - W[a^(c/b)*ln(a)/b]/ln(a)

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Now the question really is how reliable is the Lambert W function? Because you fill in your we^w=z relation and then it gives you the answer, without you knowing how they get to that answer. You can easily check it with filling your answer in we^w to check if it matches. For instance 

y = 9^x + 9x = 110

with x = 110/9 - W [ 9 ^ ( 110 / 9 ) * ln(9) / 9] / ln(9) = 2.0554779 

check: 9^2.0554779 + 9*2.0554779 ≈ 110

and you can easily check this with your calculator. But how can you get to this answer otherwise, not using the Lambert W function?

The answer again is Newton Raphson iteration:

dy/dx = (a^x) * ln(a) + a

and thus you get the iteration formula: 
xn = xi - [ a^(xi) + a*xi - b ] / [ a^(xi) * ln(a) + a ] 

for instance you guess the first answer x0 = 2.6 with a = 9 and b = 110

then you get the following iteration steps: 

x0 = 2.6

x1 = 2.279418827

x2 = 2.10129305

x3 = 2.05762191

x4 = 2.055482748

x5 = 2.055477922

As the last change is very small, then we check the f(x5) = 9^(x5) + 9*(x5) ≈ 110 (<0.0‰).

So using logarithmic differentiation conform the Newton Raphson method you can verify that the given answer by the Lambert W function is correct.

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Regarding 0^0=1, it clearly is wrong. This is an agreed upon assumption that cannot be proven.  So 0^0=1.

From my point of view if you have nothing it cannot change magically into something.

So 0^0 must be 0

In comparison 3^3=27

3^2=9 as 27/3 = 9

3^1=3 as 9/3=3

3^0=1 as 3/3=1

But 0^3=0

0^2=0

0^1=0

Then 0^0=0

As there is nothing to divide as well then it still is 0.

Conclusion: 0^0=0