What is 3ln(2e^{2x})+2ln(e^{-3x}) ?

The solution to 3ln(2e^{2x})+2ln(e^{-3x}) is 3ln(2)

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simplify 3ln(2e^{2x})+2ln(e^{-3x})

simplify

3 ln (2 e^{2 x}) + 2 ln (e^{- 3 x})

3 ln (2)

Decimal Notation

2.07944 \dots

Solution steps

3 ln (2 e^{2 x}) + 2 ln (e^{- 3 x})

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

3 ln (2 e^{2 x}) = ln ((2 e^{2 x})^{3})

= ln ((2 e^{2 x})^{3}) + 2 ln (e^{- 3 x})

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

= ln ((2 e^{2 x})^{3}) + ln ((e^{- 3 x})^{2})

Apply log rule:

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

ln ((2 e^{2 x})^{3}) + ln ((e^{- 3 x})^{2}) = ln ((e^{- 3 x})^{2} (2 e^{2 x})^{3})

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

(2 e^{2 x})^{3} (e^{- 3 x})^{2} = 8

= ln (8)

Rewrite

8

in power-base form:

8 = 2^{3}

= ln (2^{3})

Apply log rule:

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

ln (2^{3}) = 3 ln (2)

= 3 ln (2)

ln (10) + ln (7)

ln (x - 1) + ln (x + 1)

ln (\frac{sin ( x )}{1 + cos ( x )})

ln (x) - 8 [ln (x - 2) + ln (x + 2)]

\frac{1}{34} ln (\frac{11}{16})

What is 3ln(2e^{2x})+2ln(e^{-3x}) ?
The solution to 3ln(2e^{2x})+2ln(e^{-3x}) is 3ln(2)