What is ln((2e^y)/((y+1))) ?

The solution to ln((2e^y)/((y+1))) is ln(2)+y-ln(y+1)

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simplify ln((2e^y)/((y+1)))

simplify

ln (\frac{2 e ^{y}}{( y + 1 )})

ln (2) + y - ln (y + 1)

Solution steps

ln (\frac{2 e ^{y}}{( y + 1 )})

Apply log rule:

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

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

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

Apply log rule:

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

ln (2 e^{y}) = ln (2) + ln (e^{y})

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

Apply log rule:

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

ln (e^{y}) = y ln (e)

= ln (2) + y ln (e) - ln (y + 1)

y ln (e) = y

= ln (2) + y - ln (y + 1)

lo g_{10} (31.6) + 3 lo g_{10} (8 (5)) - 2.92

ln (\frac{1.26}{2 \cdot 0.085})

lo g_{10} (2 π \cdot 51782 \cdot 5.03 \cdot 9.566)

- lo g_{10} (7.5 \cdot 1 0^{- 3}) + lo g_{10} (3)

\frac{1}{2} lo g_{10} (8 + 65)

What is ln((2e^y)/((y+1))) ?
The solution to ln((2e^y)/((y+1))) is ln(2)+y-ln(y+1)