What is ln((e^x-e^{-x})/(e^{2x)(1+x^3)}) ?

The solution to ln((e^x-e^{-x})/(e^{2x)(1+x^3)}) is ln(e^x-e^{-x})-2x-ln(1+x^3)

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simplify ln((e^x-e^{-x})/(e^{2x)(1+x^3)})

simplify

ln (\frac{e ^{x} - e ^{- x}}{e ^{2 x} ( 1 + x ^{3} )})

ln (e^{x} - e^{- x}) - 2 x - ln (1 + x^{3})

Solution steps

ln (\frac{e ^{x} - e ^{- x}}{e ^{2 x} ( 1 + x ^{3} )})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{e ^{x} - e ^{- x}}{e ^{2 x} ( 1 + x ^{3} )}) = ln (e^{x} - e^{- x}) - ln (e^{2 x} (1 + x^{3}))

= ln (e^{x} - e^{- x}) - ln (e^{2 x} (x^{3} + 1))

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (e^{2 x} (1 + x^{3})) = ln (e^{2 x}) + ln (1 + x^{3})

= ln (e^{x} - e^{- x}) - (ln (e^{2 x}) + ln (x^{3} + 1))

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (e^{2 x}) = 2 x ln (e)

= ln (e^{x} - e^{- x}) - (2 x ln (e) + ln (1 + x^{3}))

2 x ln (e) = 2 x

= ln (e^{x} - e^{- x}) - (2 x + ln (x^{3} + 1))

- (2 x + ln (1 + x^{3})) : - 2 x - ln (1 + x^{3})

= ln (e^{x} - e^{- x}) - 2 x - ln (1 + x^{3})

\frac{1}{2} lo g_{2} (5) - 2 lo g_{2} (6)

6 [6 ln (x) - ln (x + 6) - ln (x - 6)]

- lo g_{10} (\frac{x}{y})

lo g_{10} (x^{3} y)

lo g_{10} (a) + 9 lo g_{10} (b)

What is ln((e^x-e^{-x})/(e^{2x)(1+x^3)}) ?
The solution to ln((e^x-e^{-x})/(e^{2x)(1+x^3)}) is ln(e^x-e^{-x})-2x-ln(1+x^3)