+0

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

0
335
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5^0/5^-6 x 1/5^-11 / (5^2)4/(5^10)2 =

Guest Jun 20, 2015

#3
+90988
+10

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 =

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

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

Melody  Jun 21, 2015
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#1
+423
+5

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:

[[5/ 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 / xy

1 / (1/x) = x

xy * xz = xy+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:

5/ 5-6*1 / 5-11 / 52*4 / 510*2

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

= 5/ 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

Sir-Emo-Chappington  Jun 20, 2015
#2
+26322
+10

.

Alan  Jun 21, 2015
#3
+90988
+10

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 =

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

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

Melody  Jun 21, 2015

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