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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

ln (z) - ln (a + 15) + ln (z - 5)

lo g_{10} (3 y (5 y + 8))

lo g_{10} ((\frac{m ^{5}}{n})^{2})

- lo g_{10} (2.3 \times 1 0^{- 8})

lo g_{10} (a) + 3 lo g_{10} (b) - 2 lo g_{10} (c)

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