What is ln((1.5x)e^{(-0.05)x}) ?

The solution to ln((1.5x)e^{(-0.05)x}) is ln(1.5)+ln(x)-0.05x

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simplify ln((1.5x)e^{(-0.05)x})

simplify

ln ((1.5 x) e^{(- 0.05) x})

ln (1.5) + ln (x) - 0.05 x

Solution steps

ln ((1.5 x) e^{(- 0.05) x})

Apply log rule:

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

ln ((1.5 x) e^{(- 0.05) x}) = ln (1.5 x) + ln (e^{(- 0.05) x})

= ln (1.5 x) + ln (e^{(- 0.05) x})

Apply log rule:

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

ln (e^{(- 0.05) x}) = (- 0.05) x ln (e)

= ln (1.5 x) + (- 0.05) x ln (e)

Apply log rule:

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

ln (1.5 x) = ln (1.5) + ln (x)

= ln (1.5) + ln (x) + (- 0.05) ln (e) x

Remove parentheses:

(- a) = - a

= ln (1.5) + ln (x) - 0.05 x ln (e)

0.05 x ln (e) = 0.05 x

= ln (1.5) + ln (x) - 0.05 x

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What is ln((1.5x)e^{(-0.05)x}) ?
The solution to ln((1.5x)e^{(-0.05)x}) is ln(1.5)+ln(x)-0.05x