What is 2ln(2xe^β)+e^β+3x ?

The solution to 2ln(2xe^β)+e^β+3x is 2ln(2)+2ln(x)+2β+e^β+3x

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simplify 2ln(2xe^β)+e^β+3x

simplify

2 ln (2 x e^{β}) + e^{β} + 3 x

2 ln (2) + 2 ln (x) + 2 β + e^{β} + 3 x

Solution steps

2 ln (2 x e^{β}) + e^{β} + 3 x

Apply log rule:

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

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

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

Apply log rule:

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

ln (e^{β}) = β ln (e)

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

2 β ln (e) = 2 β

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

\frac{5}{6} ln (5) + \frac{1}{6} ln (- 1)

ln (40) + 2 ln (\frac{1}{2}) + ln (x)

6 lo g_{10} (x) - \frac{1}{3} lo g_{10} (y)

ln (ln (\frac{1}{1 - 0.03})) - ln (ln (\frac{1}{1 - 0.008}))

ln (\frac{x}{cos ( x )})

What is 2ln(2xe^β)+e^β+3x ?
The solution to 2ln(2xe^β)+e^β+3x is 2ln(2)+2ln(x)+2β+e^β+3x