What is ln(1/m e^{(-(x-2))/m}) ?

The solution to ln(1/m e^{(-(x-2))/m}) is -(x-2)/m-ln(m)

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simplify ln(1/m e^{(-(x-2))/m})

simplify

ln (\frac{1}{m} e^{\frac{- ( x - 2 )}{m}})

- \frac{x - 2}{m} - ln (m)

Solution steps

ln (\frac{1}{m} e^{\frac{- ( x - 2 )}{m}})

Apply log rule:

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

ln (\frac{1}{m} e^{\frac{- ( x - 2 )}{m}}) = ln (\frac{1}{m}) + ln (e^{\frac{- ( x - 2 )}{m}})

= ln (\frac{1}{m}) + ln (e^{\frac{- ( x - 2 )}{m}})

Apply log rule:

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

ln (e^{\frac{- ( x - 2 )}{m}}) = \frac{- ( x - 2 )}{m} ln (e)

= ln (\frac{1}{m}) + \frac{- ( x - 2 )}{m} ln (e)

Apply log rule:

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

ln (\frac{1}{m}) = ln (1) - ln (m)

= ln (1) - ln (m) + ln (e) \frac{- ( x - 2 )}{m}

ln (1) = 0

\frac{- ( x - 2 )}{m} ln (e) = - \frac{x - 2}{m}

= 0 - ln (m) - \frac{x - 2}{m}

0 - ln (m) - \frac{x - 2}{m} = - ln (m) - \frac{x - 2}{m}

= - \frac{x - 2}{m} - ln (m)

- 6 ln (2) + \frac{15}{2} - 6 (ln (3) - ln (2)) + \frac{7}{2}

lo g_{4} (x^{2} - 11 x + 28) - lo g_{4} (x - 7)

- 3 lo g_{7} (x) + 5 lo g_{7} (y) - lo g_{3} (z)

14 + lo g_{10} (\frac{0.00084}{48})

ln (7 x^{2} - 1) - ln (x^{2})

What is ln(1/m e^{(-(x-2))/m}) ?
The solution to ln(1/m e^{(-(x-2))/m}) is -(x-2)/m-ln(m)