What is expand ln((2xe^y)^2)+e^{(y+ln(3x))} ?

The solution to expand ln((2xe^y)^2)+e^{(y+ln(3x))} is 2ln(2)+2ln(x)+2y+3e^yx

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expand ln((2xe^y)^2)+e^{(y+ln(3x))}

expand

ln ((2 x e^{y})^{2}) + e^{(y + l n (3 x))}

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

Solution steps

ln ((2 x e^{y})^{2}) + e^{(y + l n (3 x))}

Apply log rule:

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

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

= 2 ln (2 x e^{y}) + e^{y + l n (3 x)}

Apply log rule:

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

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

= 2 ln (2) + 2 ln (x) + 2 ln (e^{y}) + e^{y + l n (3 x)}

Apply log rule:

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

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

= 2 ln (2) + 2 ln (x) + 2 y ln (e) + e^{y + l n (3 x)}

2 y ln (e) = 2 y

e^{y + l n (3 x)} = 3 e^{y} x

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

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

4.41927 + 5 lo g_{10} (\frac{161}{10})

ln ∣ 13 ∣ - ln ∣ 2 ∣

ln \frac{( 1 + \frac{3}{2} )}{\frac{1}{2}}

\frac{d}{d x} (\frac{3}{x}) - ln (b x)

What is expand ln((2xe^y)^2)+e^{(y+ln(3x))} ?
The solution to expand ln((2xe^y)^2)+e^{(y+ln(3x))} is 2ln(2)+2ln(x)+2y+3e^yx