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49 to power of 7 to the power of 0

Now ${(49^{7})}^0=1$ since $x^0 =1$

However ${(49)}^{(7^0)}=(49)^1=49$.

Hence ${(49^{7})}^0 \neq {(49)}^{7^0}$

Why don't these agree?

When I see ${49^{7}}^{0}$ in questions / texts which meaning should I take?

I am aware of power towers but have not studied them before. I have a Maths Degree (UK).

amWhy
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unseen_rider
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    Because exponentiation is not associative. – John Hughes Nov 21 '17 at 18:59
  • Why shouldn't they be different? (Also ask for your money back!) – Angina Seng Nov 21 '17 at 19:02
  • Why should they agree? Nothing you learned in that Math Degree says they should agree... – David C. Ullrich Nov 21 '17 at 19:03
  • $(a^b)^c$ simplifies to $a^{bc}.$ When you see $a^{b^c}$ you should assume $a^{(b^c)}$ and not $(a^b)^c.$ And if you mean $(a^b)^c$ you should either include the parenthesis or simplify it. – Doug M Nov 21 '17 at 19:07
  • When you say, as you wrote: "49 to power of 7 to the power of 0", it is meaningless, because it is ambiguous: You list the two possible interpretations; both are legitimate, given the ambiguous statement. As you note, the interpretations result in two different answers, but we can't say which is wrong or right unless we use parentheses $(a^b)^c$ or else, $a^{(b^c)}$. From the phrase you start with, neither of this is the right one, but each is equally relevant. And so a^b^c is meaningless without parentheses. – amWhy Nov 21 '17 at 21:51

3 Answers3

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One rule for exponentiation is ${(a^b)}^c=a^{bc},$ so you have ${(49^7)}^0=49^{7*0}=49^0=1.$

When we have $a^{b^c}$, in this case, we have $49^{7^0}= 49$. These are not equal.

So the phrase "49 to power of 7 to the power of 0" is ambiguous.

As mentioned in the comments, exponentiation is not associative. The two expressions that you have have different meanings.

${(49^7)}^0$ means that you taking the product of zero $49^7$'s (that is, $\underbrace{49^7*49^7*\dotsi*49^7}_{\text{0 times}}$), so the result is the empty product, 1.

Meanwhile, $49^{(7^0)}$ means that you are taking the product of $7^0$ $49$'s. So, you have $\underbrace{49*49*\dotsi*49}_{7^0=1\text{ times}}=49.$

Air Conditioner
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  • That doesn't explain why to assume from the written statement that we mean $(a^b)^c$, instead of $a^{b^c}$? "49 to power of 7 to the power of 0" may mean $(49^7)^0$, or it may mean $49^{7^0}$. The sentence therefore is ambiguous. And you simply assumed one particular interpretation. The point is, What is your answer? You give one rule for as to how to interpret $(a^b)^c$, but fail to address what $a^{b^c}$ is nor my you chose your way over its alternative? Other than that, the answer seems to be nothing more than a (poor) summary of the comments. – amWhy Nov 21 '17 at 20:00
  • @amWhy Thank you for pointing that out. I was trying to add some information that I didn't explicitly see in previous replies that may help with OP's first question. I will delete or edit later. – Air Conditioner Nov 21 '17 at 20:07
  • No problem; I also see the asker has edited the question from the original. – amWhy Nov 21 '17 at 20:11
2

The power operation is not associative.

Note that $3^{(3^3)}=3^{27},$ whereas $(3^3)^3=3^9,$ so the phenomenon is not restricted to a case where one of the exponents is zero.

amWhy
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Mark Bennet
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There's simply no reason that $(a^b)^c$ should be the same as $a^{(b^c)}$. If you do see the notation $a^{b^c}$ you should probably assume it means $a^{(b^c)}$, because if the author meant $(a^b)^c$ he would have written $a^{bc}$ instead.

David C. Ullrich
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