What is ln(1/(2^n)*e^{-2x}e^{-3t}) ?

The solution to ln(1/(2^n)*e^{-2x}e^{-3t}) is -ln(2)n-2x-3t

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simplify ln(1/(2^n)*e^{-2x}e^{-3t})

simplify

ln (\frac{1}{2 ^{n}} \cdot e^{- 2 x} e^{- 3 t})

- ln (2) n - 2 x - 3 t

Solution steps

ln (\frac{1}{2 ^{n}} e^{- 2 x} e^{- 3 t})

Apply log rule:

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

ln (\frac{1}{2 ^{n}} e^{- 2 x} e^{- 3 t}) = ln (\frac{1}{2 ^{n}}) + ln (e^{- 2 x}) + ln (e^{- 3 t})

= ln (\frac{1}{2 ^{n}}) + ln (e^{- 2 x}) + ln (e^{- 3 t})

Apply log rule:

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

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

= ln (\frac{1}{2 ^{n}}) - 2 x ln (e) + ln (e^{- 3 t})

Apply log rule:

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

ln (e^{- 3 t}) = - 3 t ln (e)

= ln (\frac{1}{2 ^{n}}) - 2 x ln (e) - 3 t ln (e)

Apply log rule:

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

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

= ln (1) - ln (2^{n}) - 2 ln (e) x - 3 ln (e) t

Apply log rule:

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

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

= ln (1) - n ln (2) - 2 x ln (e) - 3 t ln (e)

ln (1) = 0

2 x ln (e) = 2 x

3 t ln (e) = 3 t

= 0 - ln (2) n - 2 x - 3 t

0 - n ln (2) - 2 x - 3 t = - n ln (2) - 2 x - 3 t

= - ln (2) n - 2 x - 3 t

\frac{3}{2} (lo g_{10} (4.94145) - lo g_{10} (761.21))

ln (\frac{r \cdot 0.04}{2.51})

3 lo g_{5} (a) - 3 lo g_{5} (b)

\frac{1}{2} ln (a) + 2 ln (c)

\frac{14 \cdot 0.89}{0.58} \cdot ln (\frac{1}{1 - 0.1})

What is ln(1/(2^n)*e^{-2x}e^{-3t}) ?
The solution to ln(1/(2^n)*e^{-2x}e^{-3t}) is -ln(2)n-2x-3t