Power towers and exponents

Algebra Level 1

$\large \begin{aligned} 2^2 &=& 4 \\ 2^{2^2} &=& 4\times 4 \\ 2^{2^{2^2}} &=& {4\times4\times4\times\cdots\times4} \end{aligned}$

How many 4's are there in the last equation?

5 6 7 8

4 solutions

Deva Craig
Aug 13, 2017

- $2^{2^{2^2}}$ = $2^{2^4}$

- $2^{2^4}$ = $2^{16}$

$2^{16}$ = $2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2$

$2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2$ = $4 * 4 * 4 * 4 * 4 * 4 * 4 * 4$

$2^{16}$ = $4^{8}$

Therefore, as a result, you can write $2^{2^{2^2}}$ as the product of $\boxed{8}$ number fours

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Brilliant Mathematics Staff - 3 years, 10 months ago

Sardor Yakupov
Jul 20, 2017

${ 2 }^{ { 2 }^{ { 2 }^{ 2 } } }={ 2 }^{ { 2 }^{ 4 } }={ 2 }^{ 16 }={ 4 }^{ 8 }$

Mohammad Khaza
Jul 23, 2017

${2}^{{2}^{{2}^{2}}}$ = ${2}^{{2}^{4}}$ = $2^{16}$

now, $2^{16}$ = ${2}^{{2}^{4}}$ = $4^8$

Toby M
Jul 22, 2017

By the definition of exponentiation, the number of $4$ s multiplied together in $4^k$ is $k$ .

Since $2^k = (2^2)^{k/2} = 4^{k/2}$ , we can now find $k$ and substitute it into the new expression. As $2^{2^2} = 2^4 = 16$ going from top to bottom, $2^{16} = 4^8$ . Therefore, there are $8$ $4$ s being multiplied together in $2^{2^{2^2}}$ .

Power towers and exponents

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