What is ln(1/(b^{2n)}*e^{-x/(nb)}*x) ?

The solution to ln(1/(b^{2n)}*e^{-x/(nb)}*x) is ln(x)-2nln(b)-x/(nb)

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simplify ln(1/(b^{2n)}e^{-x/(nb)}x)

simplify

ln (\frac{1}{b ^{2 n}} \cdot e^{- \frac{x}{nb}} \cdot x)

ln (x) - 2 n ln (b) - \frac{x}{nb}

Solution steps

ln (\frac{1}{b ^{2 n}} e^{- \frac{x}{nb}} x)

Apply log rule:

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

ln (\frac{1}{b ^{2 n}} e^{- \frac{x}{nb}} x) = ln (\frac{1}{b ^{2 n}}) + ln (e^{- \frac{x}{nb}}) + ln (x)

= ln (\frac{1}{b ^{2 n}}) + ln (e^{- \frac{x}{nb}}) + ln (x)

Apply log rule:

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

ln (e^{- \frac{x}{nb}}) = - \frac{x}{nb} ln (e)

= ln (\frac{1}{b ^{2 n}}) - \frac{x}{nb} ln (e) + ln (x)

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{1}{b ^{2 n}}) = ln (1) - ln (b^{2 n})

= ln (1) - ln (b^{2 n}) - ln (e) \frac{x}{nb} + ln (x)

Apply log rule:

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

ln (b^{2 n}) = 2 n ln (b)

= ln (1) - 2 n ln (b) - \frac{x}{nb} ln (e) + ln (x)

ln (1) = 0

\frac{x}{nb} ln (e) = \frac{x}{nb}

= 0 - 2 n ln (b) - \frac{x}{nb} + ln (x)

0 - 2 n ln (b) - \frac{x}{nb} + ln (x) = - 2 n ln (b) - \frac{x}{nb} + ln (x)

= ln (x) - 2 n ln (b) - \frac{x}{nb}

- \frac{1}{2} ln (\frac{0.0009}{2})

ln (1 + \frac{1}{x}) - ln (1 + \frac{1}{x - 1})

- \frac{24 ln ( 2 ) - 3 ln ( 15 )}{2 ( ln ( 15 ) - 9 ln ( 2 ) )}

lo g_{e} (\frac{100000}{1200})

2 ln (7) + ln (\frac{π ^{6}}{3})

What is ln(1/(b^{2n)}*e^{-x/(nb)}*x) ?
The solution to ln(1/(b^{2n)}*e^{-x/(nb)}*x) is ln(x)-2nln(b)-x/(nb)