Square roots

Algebra Level 2

$\Large \sqrt { \sqrt { 2 } \sqrt { 3 } \sqrt { 4 } \sqrt { 8 } \sqrt { 9 } \sqrt { 16 } \sqrt { 27 } } =2{ a }^{ { 3 }/{ 2 } }$

Find $\large a$ .

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The answer is 6.

5 solutions

Chew-Seong Cheong
Jun 1, 2015

$\begin{aligned} \sqrt{\sqrt{2}\sqrt{3}\sqrt{4}\sqrt{8}\sqrt{9} \sqrt{16}\sqrt{27}} & = \sqrt{\sqrt{2\times 3 \times 2^2 \times 2^3 \times 3^2 \times 2^4 \times 3^3}} \\ & = \sqrt{\sqrt{2^{10}3^6}} = \sqrt{2^53^3} = 2^{\frac{5}{2}}3^{\frac{3}{2}} \\ & = 2\dot{}2^{\frac{3}{2}}3^{\frac{3}{2}} = 2\dot{}6^{\frac{3}{2}} \end{aligned}$

$\Rightarrow a = \boxed{6}$

Moderator note:

Simple standard approach.

How did you separate 2 at your last step

rajesh kumar - 4 years, 5 months ago

$2\cdot 2^\frac 32 3^\frac 32 = 2(2\cdot 3)^\frac 32 = 2\cdot 6^\frac 32$ .

Chew-Seong Cheong - 4 years, 5 months ago

Michael Fuller
Jun 1, 2015

We can simplify the LHS by separating powers of $2$ and $3$ , and realising that $\sqrt { \sqrt { 16 } } =2$ , which cancels with the RHS:

$\\\large{ { a }^{ \frac { 3 }{ 2 } } =\sqrt { \sqrt { 2 } \sqrt { 4 } \sqrt { 8 } \sqrt { 3 } \sqrt { 9 } \sqrt { 27 } } \\ =\sqrt { \sqrt { { 2 }^{ 1 } } \sqrt { { 2 }^{ 2 } } \sqrt { { 2 }^{ 3 } } \sqrt { 3^{ 1 } } \sqrt { 3^{ 2 } } \sqrt { 3^{ 3 } } } \\ =\sqrt { \sqrt { { 2 }^{ 6 } } \sqrt { 3^{ 6 } } } \\ =\sqrt { { 2 }^{ 3 }\times 3^{ 3 } } \\ = \sqrt { { 6 }^{ 3 } } \\ ={ 6 }^{ \frac { 3 }{ 2 } }}\\$