Question

7. Let y = √(log3.log, 12.log, 48.log, 192 +16 - log, 12.log, 48 +10. Find y e N.​

Answer 1

SOLUTION

TO DETERMINE

$\sf{y = \sqrt{ log_{2}(3) log_{2}(12) log_{2}(48) log_{2}(192) + 16 } - log_{2}(12) log_{2}(48) + 10 }$

TO DETERMINE

The value of y ∈ N

EVALUATION

$\sf{y = \sqrt{ log_{2}(3) log_{2}(12) log_{2}(48) log_{2}(192) + 16 } - log_{2}(12) log_{2}(48) + 10 }$

Now

$\sf{log_{2}(12) }$

$\sf{= log_{2}(2 \times 2 \times 3) }$

$\sf{= log_{2}(2 \times 2 ) + log_{2}(3) }$

$\sf{= log_{2}( {2}^{2} ) + log_{2}(3) }$

$\sf{= 2 log_{2}( 2 ) + log_{2}(3) }$

$\sf{= 2+ log_{2}(3) }$

$\sf{Let \: \: x = log_{2}(3) }$

Then we have

$\sf{ log_{2}(12) = 2 + x}$

Similarly we get

$\sf{ log_{2}(48) = 4 + x}$

$\sf{ log_{2}(192) = 6 + x}$

Thus the given equation becomes

$\sf{y = \sqrt{ log_{2}(3) log_{2}(12) log_{2}(48) log_{2}(192) + 16 } - log_{2}(12) log_{2}(48) + 10 }$

$\sf{ \implies \: y = \sqrt{ x(x + 2)(x + 4)(x + 6) + 16 } - (x + 2)(x + 4) + 10 }$

$\sf{ \implies \: y = \sqrt{ x(x + 6)(x + 2)(x + 4) + 16 } - (x + 2)(x + 4) + 10 }$

$\sf{ \implies \: y = \sqrt{ ( {x}^{2} + 6x)( {x}^{2} + 6x + 8 ) + 16 } - ( {x}^{2} + 6x + 8) + 10 }$

$\sf{ let \: \: m = {x}^{2} + 6x + 8 }$

Then from above we get

$\sf{y = \sqrt{ m(m + 8)+ 16 } - ( m + 8) + 10 }$

$\sf{ \implies \: y = \sqrt{ {m}^{2} + 8m+ 16 } - ( m + 8) + 10 }$

$\sf{ \implies \: y = \sqrt{ {(m + 4)}^{2} } - ( m + 8) + 10 }$

$\sf{ \implies \: y = (m + 4) - ( m + 8) + 10 }$

$\sf{ \implies \: y = m + 4 - m - 8+ 10 }$

$\sf{ \implies \: y = 6 }$

FINAL ANSWER

Hence the required value of y = 6

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