What is e^{-ln((1/(2x^2))+c)} ?

The solution to e^{-ln((1/(2x^2))+c)} is (2x^2)/(2cx^2+1)

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simplify e^{-ln((1/(2x^2))+c)}

simplify

e^{- l n ((\frac{1}{2 x ^{2}}) + c)}

\frac{2 x ^{2}}{2 c x ^{2} + 1}

Solution steps

e^{- l n ((\frac{1}{2 x ^{2}}) + c)}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{l n (\frac{1}{2 x ^{2}} + c)}}

e^{l n (\frac{1}{2 x ^{2}} + c)} = \frac{1 + 2 c x ^{2}}{2 x ^{2}}

= \frac{1}{\frac{1 + 2 c x ^{2}}{2 x ^{2}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{2 x ^{2}}{1 + c \cdot 2 x ^{2}}

((- 1.5)^{- 4}) 6

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e^{4 x} \cdot 4 e^{4 x}

0.5 e^{- 0.1 \cdot \frac{5}{12}}

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What is e^{-ln((1/(2x^2))+c)} ?
The solution to e^{-ln((1/(2x^2))+c)} is (2x^2)/(2cx^2+1)