Math, asked by HarshalSonkusare, 6 months ago

show that log (√x^2 +1 +x) + log (√x^2 +1 -x) =0​

Answers

Answered by 66kameshtnagar
0

Answer:

log-1(-5)

Step-by-step explanation:

log 125-49('55/=:(44+'55!8")

Answered by BrainlyKingdom
0

\sf{\log _{10}\left(x+\sqrt{x^2-1}\right)+\log _{10}\left(x-\sqrt{x^2-1}\right)}

Apply Log Rule : \bf{\log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)}

\sf{=\log _{10}\left(\left(x+\sqrt{x^2-1}\right)\left(x-\sqrt{x^2-1}\right)\right)}

Apply Difference of Two Squares Formula : \bf{(a+b)(a-b)=a^2-b^2}

\sf{=\log _{10}\left(x^2-\left(\sqrt{x^2-1}\right)^2\right)}

\sf{=\log _{10}\left(x^2-\left(x^2-1\right)\right)}

\sf{=\log _{10}\left(x^2-x^2+1\right)}

\sf{=\log _{10}\left(1\right)}

Apply Log Rule : \bf{\log _a\left(1\right)=0}

\sf{=0}

Hence Proved !

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