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3^2^1

((3^2)^1) = 9

63(42^)*+5- = 63 16*+5- = 49

Solved

Posted on 2011-05-04

So I understand that converting something like:

**4 + 2 * 8 / 6 + 2**

goes from left to right, so to get the postfix you throw parentheses in it based around the precedence from left to right:

**( ( 4 + ( ( 2 * 8 ) / 6 ) ) + 2 )**

then you get the postfix by going left to right and pulling the numbers and operators based on the right parentheses:

**428*6/+2+**

But what about for exponents?

**3^2^1 **

*does this go from right to left?*

**(3^(2^1))**

*If so is this the only step thats different with exponents?*

Postfix?:

**321^^**

*Then would you evaluate this the same you would with any postfix?*

*Then does the following evaluate to 49?* For some reason I have 19 in my notes...

**6342^*+5-**

goes from left to right, so to get the postfix you throw parentheses in it based around the precedence from left to right:

then you get the postfix by going left to right and pulling the numbers and operators based on the right parentheses:

But what about for exponents?

Postfix?:

5 Comments

http://www.mathgoodies.com

3^2^1

((3^2)^1) = 9

63(42^)*+5- = 63 16*+5- = 49

http://en.wikipedia.org/wi

The TI-92 associates to the right, that is

a ^ b ^ c = a ^ (b ^ c) =

whereas, the TI-30XII associates to the left, that is

a ^ b ^ c = (a ^ b) ^ c = (ab)c.

I could have sworn that in my class the professor said it was right to left. I don't understand how something like this has no definitive rule in mathematics?

Look up in your textbook what standard you are using.

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