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$$\left\{\left[\left(\frac29\right)^4\times\left(\frac3{14}\right)^4\right]^4:\left[\left(-\frac17\right)^2\right]\right\}\times\left[\left(-\frac56\right)^3:\left(\frac5{18}\right)^3\right]^3$$

I've been trying to simplify this expression for $3$ hours. I've tried all the properties of the powers I know, and I've tried many calculators online. The result should be $-3$, but the calculators (at least the ones I've tried) and I can not calculate this expression getting the right result.

I understand it's a very simple thing, but I'm losing my mind...

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    Please see https://math.meta.stackexchange.com/questions/5020/ – Angina Seng Oct 09 '18 at 17:36
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    @LordSharktheUnknown uh thanks, sorry this is my first question. –  Oct 09 '18 at 17:39
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    What does the colon mean here? Division? (i.e., $:$ means $\div$?) Also, what value do you get when you do the calculation yourself? – Théophile Oct 09 '18 at 17:42
  • @Théophile Yes division, anyway when I calculate this I get the same result obtained by Yves Daoust –  Oct 09 '18 at 18:30
  • @J.Doe Out of curiosity, could you take a picture of the question and give a link to it? – Théophile Oct 09 '18 at 18:59
  • @Théophile Sure, https://imgur.com/a/n34sY6B –  Oct 09 '18 at 20:23
  • @J.Doe Thanks for the screenshot. It seems that you have two typos in the exponents: the first expression in square brackets should be $[\cdots]^2$, not $[\cdots]^4$; and second, you're missing the outer exponent in $\left[\left(-\frac17\right)^2\right]^4$. No wonder you were frustrated! A few minutes of careful proofreading could save you hours of headache later on. – Théophile Oct 09 '18 at 23:45
  • @Théophile wow, I checked it several times and I did not notice haha –  Oct 10 '18 at 11:46

1 Answers1

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Given the exponents, the global sign is negative.

Proceed by simplifications of the fractions (keeping an eye on the exponents).

$$\left\{\left[\left(\frac29\right)^4\times\left(\frac3{14}\right)^4\right]^4:\left[\left(\frac17\right)^2\right]\right\}\times\left[\left(\frac56\right)^3:\left(\frac5{18}\right)^3\right]^3$$

$$=\left\{\left[\left(\frac13\right)^4\times\left(\frac1{7}\right)^4\right]^4:\left[\left(\frac17\right)^2\right]\right\}\times\left[\left(\frac11\right)^3:\left(\frac1{3}\right)^3\right]^3$$

$$=\left\{\left[\left(\frac13\right)^{16}\times\left(\frac1{7}\right)^{16}\right]:\left[\left(\frac17\right)^2\right]\right\}\times\left[1:\left(\frac1{3}\right)\right]^9$$

$$=\left\{\left[\left(\frac13\right)^{7}\times\left(\frac1{7}\right)^{14}\right]\right\}.$$

Finally,

$$-\frac1{3^77^{14}}=-\frac1{1483273860320763}.$$

  • If the answer had to be $-3$, you should question the initial formula. –  Oct 09 '18 at 17:56
  • this is the initial formula, I get this from a book. –  Oct 09 '18 at 18:14
  • @J.Doe: sorry, but that's hard to believe. –  Oct 09 '18 at 18:25
  • it's a first high school book –  Oct 09 '18 at 18:28
  • @J.Doe: you'd better proofread. –  Oct 09 '18 at 18:28
  • @YvesDaoust I don't think it's hard to believe at all. Many books have typos. Yes, more often than not, it's a matter of misreading the question, but we can't summarily dismiss the possibility that maybe it's just a bad book. – Théophile Oct 09 '18 at 18:58
  • @Théophile: I'd rather trust the book than the OP. –  Oct 09 '18 at 18:59
  • @YvesDaoust That's your choice, of course, and I agree in the sense that it's more likely than not that the book is right. That being said, "hard to believe" is quite an extreme stance. I've seen books where, for example, almost every answer was wrong because the text had been updated to a new edition but not the answer key. – Théophile Oct 09 '18 at 19:07