How are exponent towers evaluated?
True or False?
For all real numbers \(a\), \(b,\) and \(c\) \[ a^{b^c} = \left(a^b\right)^c = a^{b \times c} .\]
Why some people say it's true:
The order that you choose to evaluate \(a^{b^c}\) doesn't matter, i.e. \( a^{b^c} = a^{\left(b^c\right)} = \left(a^b\right)^c = a^{b \times c}\).
Why some people say it's false:
I was taught something else.
This statement is \( \color{red} {\textbf{false}}\). Exponent towers are evaluated from the top, down.
In other words, when evaluating \( a^{b^c},\) first evaluate the top-most operation in the tower, \(b^c\), then work downward, calculating \( a^{b^c}\) as \(a^{\left(b^c\right)}\).
A simple way to remember why is that if \( a^{b^c} = \big(a^b\big) ^c = a^{bc} \), then we could have simply written it as an exponent directly in the first place. Thus, there has to be a reason why we introduce such a complicated notation, and that is because the tower is defined as \( a ^ { b^c} = a^ { ( b^c ) } \), which cannot be simplified any further.
Evaluate \[ 2^{1^3}.\]
We have \[ 2^{1^3} = 2^{1} = 2. \ _\square\]
The special cases when \(\left(a^b\right)^c = a^{(b^c)}\):
There are a few special cases when \( a^{b^c}\) does equal \(\left(a^b\right)^c\) after all. Mainly, since \(\left(a^b\right)^c = a^{b \times c}\), if \(b^c = b \times c\) then the proposed statement will be true.
The equation \(b^c = b \times c\) holds true
- whenever \(c = 1\)
- when \(b = c = 2\)
- when \(b \approx 1.495, c= 5\)
- when \(b = 4, c= 0.5\)
- ...and there are infinity many more similar examples.
Additionally, if \(a = 1 \text{ or } 0,\) then the originally proposed statement is trivially true so long as it's possible to evaluate both sides.
Rebuttal: But why is it top-down instead of bottom-up? I think bottom-up would make more sense.
Reply: You probably have a good reason for thinking that evaluating from the bottom-up is more intuitive, but that's simply not the order of operations in this case. Evaluating from the top down is also very consistent with the notion that the exponent is evaluated before all other operations, as the order "PEMDAS" defines.
Similarly to how \( 2^{3+4}\) would be evaluated as \(2^{7} = 128,\) any other operation within the power of an exponent is evaluated first, even if that operation is multiplication \(\left(\text{e.g. } 2^{3\times4} = 2^{12} = 4096\right)\) or exponentiation \(\left(\text{e.g. } 2^{3^4} = 2^{81} = \text{a very large number}\right)\).
Rebuttal: But I thought that when you have two exponents and one base, you multiply the exponents together. Doesn't \(\left(a^b\right)^c = a^{(b \times c)}\)?
Reply: With parentheses drawn in around \(a^b\), it is correct to evaluate \(a^b\) first:
\[\left(a^b\right)^c = a^{b \times c}.\]
It's only when there are no parentheses around \(a^b\) that the correct order of operations is to first evaluate \(b^c\) and then evaluate \(a^{(b^c)}\).
You might also be thinking about the multiplication \(\rightarrow\) addition identity that applies to the case when two exponents with the same base are multiplied together:
\[a^b \times a^c = a^{b+c}.\]
Want to make sure you've got this concept down? Try these problems:
\[ \LARGE 2^{2^{2^{2^0}}} = \ ? \]
\[ \LARGE 3^{2^{85}} = \, ?\]
Note: Exponent towers are evaluated as \( \large a ^ {b^c} = a^{ \left( b^ c \right) } \).
\[\huge \color{green}{2^{3^{ \left( 2^{3^{2}} \right)^{0^{2^{3}}}}}}=\color{red}{2^{3^{1^{2^{3^{2^{0^{2^{3}}}}}}}}}\]
Is the equation above true or false?
See Also