What is ln(-2/5 e^{10t}) ?

The solution to ln(-2/5 e^{10t}) is ln(2)-ln(5)+10t

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simplify ln(-2/5 e^{10t})

simplify

ln (- \frac{2}{5} e^{10 t})

ln (2) - ln (5) + 10 t

Solution steps

ln (- \frac{2}{5} e^{10 t})

Apply log rule:

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

ln (- \frac{2}{5} e^{10 t}) = ln (\frac{2}{5}) + ln (e^{10 t})

= ln (\frac{2}{5}) + ln (e^{10 t})

Apply log rule:

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

ln (e^{10 t}) = 10 t ln (e)

= ln (\frac{2}{5}) + 10 t ln (e)

Apply log rule:

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

ln (\frac{2}{5}) = ln (2) - ln (5)

= ln (2) - ln (5) + 10 ln (e) t

10 t ln (e) = 10 t

= ln (2) - ln (5) + 10 t

lo g_{e} (x^{2}) - lo g_{e} (lo g_{e} (x))

ln (\frac{e ^{x^{2}}}{x ^{2}})

- lo g_{10} (x) + \frac{1}{2} lo g_{10} (y) + 1

ln (\frac{5 g ^{- 1}}{w})

- lo g_{10} (3.55 \cdot 1 0^{- 32})