Pretty Perfect Powers

Algebra Level 3

$\underbrace{a\times a\times \cdots \times a}_{b \text{ times}} = 64$

Given the above, which of the following cannot be equal to $\underbrace{b\times b\times \cdots \times b}_{a \text{ times}}?$

1\times 1\times 1 \times 1

2\times 2\times 2 \times 2

3\times 3\times 3 \times 3

4\times 4\times 4 \times 4

1 solution

Chandrashekar Giridharan
May 16, 2018

64 can be represented as $64^1,8^2 ,4^3$ . The values we are looking for are obtained by basically switches the values of the base and the exponent. That is, if $64=a^b$ ,then we are looking for the value $b^a$

Therefore we can get $1^{64}=1\times1\times1\times1$

$2^8=4\times4\times4\times4$

$3^4=3\times3\times3\times3$

Now we can say that the only value we cannot get is $2\times2\times2\times2$

I think it's better to clarify more with saying that a and b are integers.

Mustafa Alelg - 3 years ago

No.

If $a$ and $b$ are not integers, or positive, then the expressions $\underbrace{a\times a\times \cdots \times a}_{b \text{ times}}$ and $\underbrace{b\times b\times \cdots \times b}_{a \text{ times}}$ are meaningless.

Pi Han Goh - 3 years ago

Isn't this just the definition of the exponents? if it's so then it's not meaningless.

Mustafa Alelg - 3 years ago

@Mustafa Alelg – There's a subtle difference here, the value of $b$ in $a^b$ is not necessarily an integer, but the value of $b$ in $\underbrace{a\times a\times \cdots \times a}_{b \text{ times}}$ must be both positive and an integer.

Pi Han Goh - 3 years ago

@Pi Han Goh – Now I understand.. thank you!

Mustafa Alelg - 3 years ago

Pretty Perfect Powers

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