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Now, it may sound absurd, but I would like to know (if exists), how to read $(x^{n-1})$ and $(x^n-1)$ distinctly and practice this it in everyday Mathematics.

As of now, I read $(x^{n-1})$ as, "x whole raised to power n-1" and $(x^n-1)$ as " x raised to n whole minus 1".

Similar confusing scenarios are also welcome.

Thank you.

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    $\begingroup$ Maybe "x to the nth power minus 1" and "x to the n minus one-th power"? $\endgroup$ – angryavian Dec 31 '20 at 3:22
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    $\begingroup$ I typically read this using pauses. The first I read as "x to the n minus one" whereas the second I would read as "x to the n minus one". $\endgroup$ – K.defaoite Dec 31 '20 at 3:28
  • $\begingroup$ Sometimes I say "$x$ to the quantity $n-1$" for $x^{n-1}$. $\endgroup$ – littleO Dec 31 '20 at 3:38
  • $\begingroup$ I once sat on a committee in which the candidates had to take a written Math exam and then read aloud everything they wrote in public, including formulas and computations. I guess I don't need to say that nobody understood anything. $\endgroup$ – Ruy Dec 31 '20 at 4:34
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To read aloud I would say, "x to the power, pause, n minus 1", and "x to the power n, pause, minus 1". The position of the pause is used to communicate the implied brackets. If this is not clear enough, then "x to the power open bracket n minus 1 close bracket".

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For $x^{n-1}$, $``x$ whole raised to $n-1"$ looks great.

For $x^n - 1$, I'd suggest $``1$ less than $x^n"$

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  • $\begingroup$ The second definition suggests $1-x^n=x^n-1$, which is clearly false. You should write $-1+x^n$. $\endgroup$ – manooooh Dec 31 '20 at 3:26
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    $\begingroup$ No, It's not @manooooh. $\endgroup$ – emil Dec 31 '20 at 3:28
  • $\begingroup$ Isn't "less" the same as "minus" in these contexts, @emil? $\endgroup$ – manooooh Dec 31 '20 at 3:29
  • $\begingroup$ @manooooh No, it's not! $\endgroup$ – strawberry-sunshine Dec 31 '20 at 3:40
  • $\begingroup$ maybe “less” but not “less than” $\endgroup$ – J. W. Tanner Dec 31 '20 at 3:41
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For $x^{n-1}$: $x$ raised to the difference of $n$ and $1$. And for $x^n-1$: The difference of $x$, raised to $n$, and $1$.

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