Log -e2 - 1 - 1-e2 - 1 - 1

log (√(e^2+1 1)/√(e^2+1+1))= log(√(e^2)/√(e^2+2)) = log(e/√(e^2+2)) #160; #160; [∵ log (a/b)=loga logb] = log e log(e^2+2)^1/2 ...

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Byju's Answer

Standard XII

Mathematics

Special Integrals - 1

log √e2 + 1 -...

Question

$log (\frac{\sqrt{e^{2} + 1 - 1}}{\sqrt{e^{2} + 1 + 1}})$

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Solution

$l o g (\frac{\sqrt{e^{2} + 1 - 1}}{\sqrt{e^{2} + 1 + 1}}) = l o g (\frac{\sqrt{e^{2}}}{\sqrt{e^{2} + 2}})$

$= l o g (\frac{e}{\sqrt{e^{2} + 2}})$ $[∵ l o g (\frac{a}{b}) = l o g a - l o g b]$

$= l o g e - l o g (e^{2} + 2)^{1 / 2}$

$= l o g e - \frac{1}{2} l o g (e^{2} + 2)$

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log(√e2+1−1√e2+1+1)

log(√e2+1−1√e2+1+1)=log(√e2√e2+2)=log(e√e2+2) [∵log(ab)=loga−logb]=loge−log(e2+2)1/2=loge−12log(e2+2)

$log (\frac{\sqrt{e^{2} + 1 - 1}}{\sqrt{e^{2} + 1 + 1}})$

$l o g (\frac{\sqrt{e^{2} + 1 - 1}}{\sqrt{e^{2} + 1 + 1}}) = l o g (\frac{\sqrt{e^{2}}}{\sqrt{e^{2} + 2}})$

$= l o g (\frac{e}{\sqrt{e^{2} + 2}})$ $[∵ l o g (\frac{a}{b}) = l o g a - l o g b]$

$= l o g e - l o g (e^{2} + 2)^{1 / 2}$

$= l o g e - \frac{1}{2} l o g (e^{2} + 2)$