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# Index laws question... only one that stumped me!

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173
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I need some help simplifying 'e'.

Thanks for any help!

Oct 4, 2018

#1
+4833
+2

$$\Large \dfrac{a^{2n-1}\times b^3 \times c^{1-n}}{a^{n-3}\times b^{2-n}\times c^{2-2n}} = \\ \\ \Large a^{(2n-1)-(n-3))}\times b^{3-(2-n)}\times c^{(1-n)-(2-2n)} = \\ \\ \Large a^{n+2}\times b^{n+1}\times c^{n-1}$$

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Oct 4, 2018

#1
+4833
+2

$$\Large \dfrac{a^{2n-1}\times b^3 \times c^{1-n}}{a^{n-3}\times b^{2-n}\times c^{2-2n}} = \\ \\ \Large a^{(2n-1)-(n-3))}\times b^{3-(2-n)}\times c^{(1-n)-(2-2n)} = \\ \\ \Large a^{n+2}\times b^{n+1}\times c^{n-1}$$

Rom Oct 4, 2018
#2
+52
-2

Ok. 'e' is Euler's Constant, or approximately 2.718281828459045235360...

Euler's constant can be defined by terms of an infinite sum in the form of $$e=\sum_{n=0}^{}{\frac{1}{n!}}$$

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Oct 4, 2018
#3
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no thats lord farquaad from shrek

Guest Oct 4, 2018