What is ln(a(2x+1)e^{-ax(x+1)}) ?

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

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simplify ln(a(2x+1)e^{-ax(x+1)})

simplify

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

ln (a) + ln (2 x + 1) - a x (x + 1)

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

= ln (a) + ln (2 x + 1) - a x (x + 1) ln (e)

a x (x + 1) ln (e) = a x (x + 1)

= ln (a) + ln (2 x + 1) - a x (x + 1)

\frac{2 ln ( 2 )}{ln ( 5 )} - 4

lo g_{10} (y^{5} x)

ln (- sin (x) + 1) - ln (sin (x) + 1)

ln (\frac{35.35 - 0}{18.25 - 0})

ln (\frac{1}{1 - v})

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