Question

If 1003x + 1003-X = 4, then the value of √1003 8x-1003-8x/679(1003-1003-x)​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ {1003}^{x} + {1003}^{ - x} = 4 }$

TO DETERMINE

$\displaystyle \sf{ \sqrt{ \frac{{1003}^{8x} - {1003}^{ - 8x}}{679({1003}^{x} - {1003}^{ - x})} } }$

EVALUATION

Here it is given that

$\displaystyle \sf{ {1003}^{x} + {1003}^{ - x} = 4 }$

$\displaystyle \sf{ Let \: \: \: y = {1003}^{x}}$

Then above becomes

$\boxed{ \displaystyle \sf{y + \frac{1}{y} = 4 }}$

Squaring both sides we get

$\displaystyle \sf{ {y}^{2} + \frac{1}{ {y}^{2} } + 2 = 16 }$

$\boxed{ \: \displaystyle \sf{ \implies \: {y}^{2} + \frac{1}{ {y}^{2} } = 14 } \: }$

Again Squaring both sides we get

$\displaystyle \sf{ \implies \: {y}^{4} + \frac{1}{ {y}^{4} } + 2 = 196}$

$\boxed{ \: \displaystyle \sf{ \implies \: {y}^{4} + \frac{1}{ {y}^{4} } = 194} \: \: }$

Now

$\displaystyle \sf{ \sqrt{ \frac{{1003}^{8x} - {1003}^{ - 8x}}{679({1003}^{x} - {1003}^{ - x})} } }$

$\displaystyle \sf{ = \sqrt{ \frac{{y}^{8} - \frac{1}{ {y}^{8} } }{679 \bigg(y - \frac{1}{y} \bigg) } } }$

$\displaystyle \sf{ = \sqrt{ \frac{ \bigg({y}^{4} + \frac{1}{ {y}^{4} } \bigg)\bigg({y}^{4} - \frac{1}{ {y}^{4} } \bigg) }{679 \bigg(y - \frac{1}{y} \bigg) } } }$

$\displaystyle \sf{ = \sqrt{ \frac{ \bigg({y}^{4} + \frac{1}{ {y}^{4} } \bigg)\bigg({y}^{2} + \frac{1}{ {y}^{2} } \bigg)\bigg({y}^{2} - \frac{1}{ {y}^{2} } \bigg) }{679 \bigg(y - \frac{1}{y} \bigg) } } }$

$\displaystyle \sf{ = \sqrt{ \frac{ \bigg({y}^{4} + \frac{1}{ {y}^{4} } \bigg)\bigg({y}^{2} + \frac{1}{ {y}^{2} } \bigg) \bigg(y + \frac{1}{y} \bigg)\bigg(y - \frac{1}{y} \bigg) }{679 \bigg(y - \frac{1}{y} \bigg) } } }$

$\displaystyle \sf{ = \sqrt{ \frac{ \bigg({y}^{4} + \frac{1}{ {y}^{4} } \bigg)\bigg({y}^{2} + \frac{1}{ {y}^{2} } \bigg) \bigg(y + \frac{1}{y} \bigg)}{679 } } }$

$\displaystyle \sf{ = \sqrt{ \frac{194 \times 14 \times 4}{679} } }$

$\displaystyle \sf{ = \sqrt{ \frac{2 \times 14 \times 4}{7} } }$

$\displaystyle \sf{ = \sqrt{ 2 \times 2 \times 4} }$

$\displaystyle \sf{ = 4}$

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