5^0/5^-6 x 1/5^-11 / (5^2)4/(5^10)2 =

Thanks Sir-Emo-Chappington and Alan   

We could get  a job writing multiple choice answers.

We only have 3 answers so far - Someone should offer a 4th.         Any takers    

5^0/5^-6 x 1/5^-11 / (5^2)4/(5^10)2 =

$$\\\frac{5^0}{5^{-6}} \times 1/(5^{-11}) / (5^2)\times 4/(5^{10})\times 2 \\\\
=1*5^6 /(5^{-11}) / (5^2)\times 4/(5^{10})\times 2 \\\\
=(5^6) *(5^{11}) / (5^2)\times 4/(5^{10})\times 2 \\\\
=\frac{(5^6) *(5^{11}) }{(5^2)}\times 4/(5^{10})\times 2 \\\\
=\frac{4*(5^6) *(5^{11}) }{(5^2)}/(5^{10})\times 2 \\\\
=\frac{4*(5^6) *(5^{11}) }{(5^2)*(5^{10})}\times 2 \\\\
=\frac{2*4*(5^6) *(5^{11}) }{(5^2)*(5^{10})} \\\\
=\frac{8*(5^{17}) }{(5^{12})} \\\\
=8*(5^{5}) } \\\\$$

$${\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{5}}} = {\mathtt{25\,000}}$$

There's quite a lot of different ways of interpretting this question, so apologies if I got this incorrect.

I ordered it around in the way that looked most logical to me:

[[5⁰ / 5^-6]*[1 / 5^-11]] / [[(5²)*4 / (5¹⁰)*2]]

$${\frac{\left[{\frac{{{\mathtt{5}}}^{{\mathtt{0}}}}{{{\mathtt{5}}}^{-{\mathtt{6}}}}}\right]{\mathtt{\,\times\,}}\left[{\frac{{\mathtt{1}}}{{{\mathtt{5}}}^{-{\mathtt{11}}}}}\right]}{\left[{\frac{\left({{\mathtt{5}}}^{{\mathtt{2}}}\right){\mathtt{\,\times\,}}{\mathtt{4}}}{\left[\left({{\mathtt{5}}}^{{\mathtt{10}}}\right){\mathtt{\,\times\,}}{\mathtt{2}}\right]}}\right]}}$$

Some quick formulae to explain my workings:

x⁰ = 1

x^-y = 1 / x^y

1 / (1/x) = x

x^y * x^z = x^y+z

So this equation is = [(1 / (1/5⁶⁾) * (1 / (1/5¹¹))] / [(5²*4) / (5¹⁰*2)]

= [5⁶ * 5¹¹] / [(5²*4) / (5¹⁰*2)]

= 5¹⁷ / [(5²*4) / (5¹⁰*2)]

= 762939453125 / [(25*4) / 9765625*2]

= 762939453125 / 0.00000512

= 149011611938476562.5

= 1.49 * 10¹⁷

Of course I could read it without assuming any of what I did, and just work on the basic principles of what order you should do each function (however even then I must assume where each power "ends", since otherwise you have a crazy looking equation), in which case I get:

5⁰ / 5^-6*1 / 5^-11 / 5²*4 / 5¹⁰*2

$${\frac{\left({\frac{\left({\frac{\left({\frac{{{\mathtt{5}}}^{{\mathtt{0}}}}{\left({{\mathtt{5}}}^{{\mathtt{6}}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}}\right)}{{{\mathtt{5}}}^{{\mathtt{11}}}}}\right)}{\left(\left({{\mathtt{5}}}^{{\mathtt{2}}}\right){\mathtt{\,\times\,}}{\mathtt{4}}\right)}}\right)}{\left(\left({{\mathtt{5}}}^{{\mathtt{10}}}\right){\mathtt{\,\times\,}}{\mathtt{2}}\right)}}$$

= 5⁰ / 5^-6 / 5^-11 / 5²*4 / 5¹⁰*2

= (((1 / 0.000064) / 0.00000002048) / 100) / 19531250

= ((15625 / 0.00000002048) / 100) / 19531250

= (762939453125 / 100) / 19531250

= 7629394531.25 / 19531250

= 390.625

If I truly do not assume anything ends, then what I get is:

$${{\mathtt{5}}}^{\left({\frac{{\mathtt{0}}}{{{\mathtt{5}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{1}}}{{{\mathtt{5}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{11}}}{\left({\frac{{{\mathtt{5}}}^{\left({\mathtt{2}}\right)}{\mathtt{\,\times\,}}{\mathtt{4}}}{\left({{\mathtt{5}}}^{\left({\mathtt{10}}\right)}{\mathtt{\,\times\,}}{\mathtt{2}}\right)}}\right)}}\right)}}}\right)}}}\right)}$$

Which has a fraction with 0 on top, so in short ends up as 5⁰ = 1

.