What is ln(e^v(p^dq^{1-d})) ?

The solution to ln(e^v(p^dq^{1-d})) is v+dln(p)+ln(q)(1-d)

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simplify ln(e^v(p^dq^{1-d}))

simplify

ln (e^{v} (p^{d} q^{1 - d}))

v + d ln (p) + ln (q) (1 - d)

Solution steps

ln (e^{v} (p^{d} q^{1 - d}))

Apply log rule:

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

ln (e^{v} (p^{d} q^{1 - d})) = ln (e^{v}) + ln (p^{d} q^{1 - d})

= ln (e^{v}) + ln (p^{d} q^{- d + 1})

Apply log rule:

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

ln (e^{v}) = v ln (e)

= v ln (e) + ln (p^{d} q^{1 - d})

Apply log rule:

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

ln (p^{d} q^{1 - d}) = ln (p^{d}) + ln (q^{1 - d})

= ln (e) v + ln (p^{d}) + ln (q^{- d + 1})

Apply log rule:

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

ln (p^{d}) = d ln (p)

= v ln (e) + d ln (p) + ln (q^{1 - d})

Apply log rule:

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

ln (q^{1 - d}) = (1 - d) ln (q)

= v ln (e) + d ln (p) + (1 - d) ln (q)

v ln (e) = v

= v + d ln (p) + ln (q) (- d + 1)

ln (\frac{9}{5}) - \frac{16}{45}

ln (\frac{1000}{25})

- 10 ln (\frac{1}{x + 1})

ln (\frac{150}{625})

ln (3 (3))

What is ln(e^v(p^dq^{1-d})) ?
The solution to ln(e^v(p^dq^{1-d})) is v+dln(p)+ln(q)(1-d)