2 \sqrt2 is 2 \sqrt2

Algebra Level 2

Find the value of:

\LARGE { { \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \left( { \sqrt { 2 } }^{ \sqrt { 2 } } \right) ^\sqrt2 } } } } }

Try more Number Theory Problems

2 \sqrt2 Can't be ditermined 1 \infty 2 0

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1 solution

Md Mehedi Hasan
Nov 7, 2017

\LARGE { { \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \left( { \sqrt { 2 } }^{ \sqrt { 2 } } \right) ^\sqrt2 } } } } ={ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { { \sqrt { 2 } }^2 } } } } ={ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \sqrt { 2 } }^2 } } ={ \sqrt { 2 } }^{ { \sqrt { 2 } }^2 } ={ \sqrt { 2 } }^2 =\boxed2}

How is . 2 2 2 2 = 2 2 2 2 2 .{ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \sqrt { 2 } }^2 } } ={ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { { \sqrt { 2 } }^2 } } } }

Sumukh Bansal - 3 years, 7 months ago

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I can't understand, what do you want to say. Please clearly say....

Md Mehedi Hasan - 3 years, 7 months ago

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I am unable to understand the second and third steps of your solving.

Sumukh Bansal - 3 years, 7 months ago

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@Sumukh Bansal I see...

We know ( a m ) n = a m n \left(a^m\right)^n=a^{mn}

For this in 1st step, I write ( 2 2 ) 2 = 2 ( 2 × 2 ) = 2 2 = 2 \left({\sqrt2}^{\sqrt2}\right)^{\sqrt2}={\sqrt2}^{\left(\sqrt2\times\sqrt2\right)}={\sqrt2}^2=2

In 2nd step, I write 2 2 = ( 2 ) 2 = 2 {\sqrt2}^2=\left(\sqrt2\right)^2=2

And continuously go on...

Md Mehedi Hasan - 3 years, 7 months ago

@Sumukh Bansal Can you realize now? Anything else, you can ask me...

Md Mehedi Hasan - 3 years, 7 months ago

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@Md Mehedi Hasan Thanks for your help @Md Mehedi Hasan

Sumukh Bansal - 3 years, 7 months ago

@Md Mehedi Hasan Is this problem and solution correct

Sumukh Bansal - 3 years, 7 months ago

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