${(49^{7})}^0=1$ however ${(49)}^{(7^0)}=49$ why don't they equal?

Question

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).

Answer 1

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 $\underbrace{49*49*\dotsi*49}_{7^0=1\text{ times}}=49.$

Answer 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.

Answer 3

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.