log-10-x-2-1-2-log-10-x-2-1-2-

Question Number 4478 by love math last updated on 30/Jan/16

{l o g}_{10} (x^{2} + 1) = \frac{2}{{l o g}_{10} (x^{2} + 1)} - 2

Answered by Rasheed Soomro last updated on 30/Jan/16

{[{l o g}_{10} (x^{2} + 1)]}^{2} = 2 - 2 ({l o g}_{10} (x^{2} + 1))

{[{l o g}_{10} (x^{2} + 1)]}^{2} + 2 ({l o g}_{10} (x^{2} + 1)) - 2 = 0

l e t {l o g}_{10} (x^{2} + 1) = y

y^{2} + 2 y - 2 = 0

y = \frac{- 2 \pm \sqrt{2^{2} - 4 (1) (- 2)}}{2 (1)}

y = \frac{- 2 \pm \sqrt{12}}{2} = - 1 \pm \sqrt{3}

{l o g}_{10} (x^{2} + 1) = - 1 \pm \sqrt{3}

x^{2} + 1 = 10^{- 1 \pm \sqrt{3}}

x = \pm \sqrt{10^{- 1 \pm \sqrt{3}} - 1}

{\sqrt{10^{- 1 + \sqrt{3}} - 1}, \sqrt{10^{- 1 - \sqrt{3}} - 1}, - \sqrt{10^{- 1 + \sqrt{3}} - 1}, \sqrt{10^{- 1 - \sqrt{3}} - 1}}

Commented by RasheedSindhi last updated on 30/Jan/16

C o n t i n u e f r o m a n s w e r

y = - 1 \pm \sqrt{3}

x^{2} + 1 = - 1 \pm \sqrt{3}

x^{2} = - 2 \pm \sqrt{3}

x = \pm \sqrt{- 2 + \sqrt{3}}, \pm \sqrt{- 2 - \sqrt{3}}

A l l r o o t s a r e c o m p l e x, b e c a u s e

n u m b e r s u n d e r \sqrt{} a r e n e g a t i v e .

Commented by FilupSmith last updated on 31/Jan/16

You wrote above :

x^{2} + 1 = - 1 \pm \sqrt{3}

Is this correct ?

It was said that y = \log_{10} (x^{2} + 1)

∴ \log_{10} (x^{2} + 1) = - 1 \pm \sqrt{3}

x^{2} = 10^{- 1 \pm \sqrt{3}} - 1

∴ x = \pm \sqrt{10^{- (1 - \sqrt{3})} - 1}, \pm \sqrt{10^{- (1 + \sqrt{3})} - 1}

Have I misread your working ?

Commented by RasheedSindhi last updated on 30/Jan/16

T h a n k s! {I t}^{'} s a m i s t a k e .

Y o u a r e v e r y r i g h t! I a m g o i n g t o

c o r r e c t m y a n s w e r .