Note that if the equation really is (in traditional notation)

4^x - 2^x - 8 = 0

then it can be rewritten as

y^2 - y - 8 = 0, y = 2^x

and solved in closed form as well,
yielding a countably infinite set of solutions aligned along one (or two) 
vertical lines in the complex plane.
(If I am not mistaken!)

Louis

> On 29 Jan 2018, at 07:19, Rob Hodgkinson <[email protected]> wrote:
> 
> @Skip et al …
> 
> also apologies for my sill definitions, I should have used y inside the 
> definitions not x (!!!), sorry if I confused the issue…
> 
> as in here for the first interpretation …
> 
> x,"0 (3 : '8+(2^y)-((2^2)^y)') x=:1.6+0.05*i.8
> 
> …/Rob
> 
>> On 29 Jan 2018, at 3:49 pm, Jose Mario Quintana 
>> <[email protected]> wrote:
>> 
>> In that case,
>> 
>>  (X=. (8 + (2 ^ ]) - (2 ^ 2) ^ ])New (^:22) 1)
>> 1.75372489
>> 
>> is a root,
>> 
>>  (8 + (2 ^ ]) - (2 ^ 2) ^ ])X
>> 0
>> 
>> but, it is not the only one,
>> 
>>  (X=. (8 + (2 ^ ]) - (2 ^ 2) ^ ])New (^:22) 0.5j_0.5)
>> 1.24627511j4.53236014
>> 
>>  (8 + (2 ^ ]) - (2 ^ 2) ^ ])X
>> 8.8817842e_16j7.72083702e_15
>> 
>> 
>> 
>>> On Sun, Jan 28, 2018 at 9:27 PM, Raul Miller <[email protected]> wrote:
>>> 
>>> Hmm...
>>> 
>>> I had originally thought about calling out the (2^2)^x interpretation
>>> as a possibility, because rejected that, because that would be better
>>> expressed as 4^x
>>> 
>>> But it's possible that Skip got the 1.75379 number from someone who
>>> thought different about this.
>>> 
>>> And, to be honest, it is an ambiguity in the original expression -
>>> just one that I thought should be rejected outright, rather than
>>> suggested.
>>> 
>>> Which gets us into another issue, which is that what one person would
>>> think is obviously right is almost always what some other person would
>>> think is obviously wrong... (and this issue crops up all over, not
>>> just in mathematic and/or programming contexts).
>>> 
>>> --
>>> Raul
>>> 
>>> 
>>>> On Sun, Jan 28, 2018 at 9:08 PM, Rob Hodgkinson <[email protected]> wrote:
>>>> @Skip
>>>> 
>>>> Skip, I am a confused in your original post… your actual post read;
>>>> 
>>>> ================================================
>>>> What is the best iterative way to solve this equation:
>>>> (-2^2^x) + (2^x) +8 =0
>>>> then later to Raul,
>>>> 0 = 8 + (2^x) - 2^2^x    NB. Is correct, and the answer is real
>>>> The answer is close to 1.75379
>>>> ================================================
>>>> 
>>>> However I suspect your original syntax was not J syntax (could it have
>>> been math type syntax ?) as it differs on the J style right-to-left syntax
>>> on the 2^2^x expression.
>>>> 
>>>> The 2 possible interpretations are shown below and only the Excel type
>>> syntax seems to get close to your expected answer.
>>>> 
>>>> NB. Excel interpretation
>>>>  x,"0 (3 : '8+(2^x)-((2^2)^x)') x=:1.6+0.05*i.8
>>>> 1.6        1.84185
>>>> 1.65       1.28918
>>>> 1.7        0.692946
>>>> 1.75       0.0498772                NB. intercept seems close to your
>>> expected value of 1.75379
>>>> 1.8    _0.64353
>>>> 1.85  _1.39104
>>>> 1.9    _2.19668
>>>> 1.95  _3.06478
>>>> 
>>>> NB. J style interpretation
>>>>  x,"0 (3 : '8+(2^x)-(2^2^x)') x=:1.6+0.05*i.8
>>>> 1.6      2.85522
>>>> 1.65    2.33325
>>>> 1.7      1.74188
>>>> 1.75    1.07063
>>>> 1.8      0.307207                 NB. But in this model the intercept is
>>> above 1.8, but this is the model that has been coded in responses to your
>>> post ??
>>>> 1.85  _0.562844
>>>> 1.9    _1.5566
>>>> 1.95  _2.69431
>>>> 
>>>> Please clarify, thanks, Rob
>>>> 
>>>> 
>>>>> On 29 Jan 2018, at 12:48 pm, Jose Mario Quintana <
>>> [email protected]> wrote:
>>>>> 
>>>>> Moreover, apparently there is at least another solution,
>>>>> 
>>>>> ((-2^2^X) + (2^X) +8 ) [  X=. 2.9992934709539156j_13.597080425481581
>>>>> 5.19549681e_16j_2.92973749e_15
>>>>> 
>>>>> 
>>>>> On Sun, Jan 28, 2018 at 7:28 PM, Jose Mario Quintana <
>>>>> [email protected]> wrote:
>>>>> 
>>>>>> Are you sure?
>>>>>> 
>>>>>> u New
>>>>>> - (u %. u D.1)
>>>>>> 
>>>>>> ,. (8 + (2 ^ ]) - 2 ^ 2 ^ ])New (^:(<22)) 1
>>>>>>       1
>>>>>> 3.44167448
>>>>>> 3.25190632
>>>>>> 3.03819348
>>>>>> 2.7974808
>>>>>> 2.53114635
>>>>>> 2.25407823
>>>>>> 2.00897742
>>>>>> 1.86069674
>>>>>> 1.82070294
>>>>>> 1.81842281
>>>>>> 1.81841595
>>>>>> 1.81841595
>>>>>> 1.81841595
>>>>>> 1.81841595
>>>>>> 1.81841595
>>>>>> 1.81841595
>>>>>> 1.81841595
>>>>>> 1.81841595
>>>>>> 1.81841595
>>>>>> 1.81841595
>>>>>> 1.81841595
>>>>>> 
>>>>>> (8 + (2 ^ ]) - 2 ^ 2 ^ ]) 1.81841595
>>>>>> _8.21739086e_8
>>>>>> 
>>>>>> (8 + (2 ^ ]) - 2 ^ 2 ^ ]) 1.75379
>>>>>> 1.01615682
>>>>>> 
>>>>>> PS.  Notice that New is a tacit version (not shown) of Louis' VN
>>> adverb.
>>>>>> 
>>>>>> 
>>>>>> 
>>>>>> 
>>>>>> On Sun, Jan 28, 2018 at 5:52 PM, Skip Cave <[email protected]>
>>>>>> wrote:
>>>>>> 
>>>>>>> Raul,
>>>>>>> 
>>>>>>> You had it right in the first place.
>>>>>>> 
>>>>>>> 0 = 8 + (2^x) - 2^2^x    NB. Is correct, and the answer is real
>>>>>>> 
>>>>>>> The answer is close to 1.75379
>>>>>>> 
>>>>>>> I wanted to know how to construct the Newton Raphson method using the
>>>>>>> iteration verb N described in the link: http://code.jsoftware.
>>>>>>> com/wiki/NYCJUG/2010-11-09
>>>>>>> under "A Sampling of Solvers - Newton's Method"
>>>>>>> 
>>>>>>> N=: 1 : '- u % u d. 1'
>>>>>>> 
>>>>>>> Skip
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>> Skip Cave
>>>>>>> Cave Consulting LLC
>>>>>>> 
>>>>>>> On Sun, Jan 28, 2018 at 4:38 PM, Raul Miller <[email protected]>
>>>>>>> wrote:
>>>>>>> 
>>>>>>>> Eh... I *think* you meant what would be expressed in J as:
>>>>>>>> 
>>>>>>>> 0 = 8 + (2^x) - 2^2^x
>>>>>>>> 
>>>>>>>> I'd probably try maybe a few hundred rounds of newton's method first,
>>>>>>>> and see where that leads.
>>>>>>>> 
>>>>>>>> But there's an ambiguity where the original expression (depending on
>>>>>>>> the frame of reference of the poster) could have been intended to be:
>>>>>>>> 
>>>>>>>> 0 = 8 + (2^x) + _2^2^x
>>>>>>>> 
>>>>>>>> [if that is solvable, x might have to be complex]
>>>>>>>> 
>>>>>>>> Thanks,
>>>>>>>> 
>>>>>>>> --
>>>>>>>> Raul
>>>>>>>> 
>>>>>>>> On Sun, Jan 28, 2018 at 5:25 PM, Skip Cave <[email protected]>
>>>>>>>> wrote:
>>>>>>>>> What is the best iterative way to solve this equation:
>>>>>>>>> 
>>>>>>>>> (-2^2^x) + (2^x) +8 =0
>>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> Skip Cave
>>>>>>>>> Cave Consulting LLC
>>>>>>>>> ------------------------------------------------------------
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>>>>>> 
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