What is ln(e^{2x+y}*(e^y-1)) ?

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

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simplify ln(e^{2x+y}*(e^y-1))

simplify

ln (e^{2 x + y} \cdot (e^{y} - 1))

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

(2 x + y) ln (e) = 2 x + y

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

ln (10) - ln (12)

- ln (\frac{10}{20})

ln (\frac{ln ( x ^{2} )}{4 x})

ln (\frac{( 48 ) - ( 9 )}{( 70 ) - ( 9 )})

ln (\frac{658405861}{2405861})

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