What is expand ln(x)(e^xv_{1}+(e^x)/x v_{2}=0) ?

The solution to expand ln(x)(e^xv_{1}+(e^x)/x v_{2}=0) is e^xv_{1}ln(x)+(e^xv_{2}ln(x))/x =0

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expand ln(x)(e^xv_{1}+(e^x)/x v_{2}=0)

expand

ln (x) (e^{x} v_{1} + \frac{e ^{x}}{x} v_{2} = 0)

e^{x} v_{1} ln (x) + \frac{e ^{x} v _{2} ln ( x )}{x} = 0

Solution steps

ln (x) (e^{x} v_{1} + \frac{e ^{x}}{x} v_{2} = 0)

= ln (x) (e^{x} v_{1} + 0 = \frac{e ^{x}}{x} v_{2})

Apply the distributive law:

a (b + c) = ab + a c

a = ln (x), b = e^{x} v_{1}, c = \frac{e ^{x}}{x} v_{2} = 0

ln (x) e^{x} v_{1} + ln (x) \frac{e ^{x}}{x} v_{2} = 0

e^{x} v_{1} ln (x) + \frac{e ^{x}}{x} v_{2} ln (x) = 0

Expand

e^{x} v_{1} ln (x) + \frac{e ^{x}}{x} v_{2} ln (x) : e^{x} v_{1} ln (x) + \frac{e ^{x} v _{2} ln ( x )}{x}

e^{x} v_{1} ln (x) + \frac{e ^{x} v _{2} ln ( x )}{x} = 0

(- 2 y + 8 + 2)^{2}

(4 y - 5 i)^{2}

\frac{e ^{- 2 s}}{( s ^{2} + 9 ) ( s ^{2} + 1 )}

(k + \frac{1}{3}) \cdot (k - \frac{2}{3})

C (0, - 3) yD (- 1, 7)

What is expand ln(x)(e^xv_{1}+(e^x)/x v_{2}=0) ?
The solution to expand ln(x)(e^xv_{1}+(e^x)/x v_{2}=0) is e^xv_{1}ln(x)+(e^xv_{2}ln(x))/x =0