What is 1/2 ln(e^{mx}e^{my}(1-z^2)^{m/2}) ?

The solution to 1/2 ln(e^{mx}e^{my}(1-z^2)^{m/2}) is (mx+my)/2+(mln(1-z^2))/4

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simplify 1/2 ln(e^{mx}e^{my}(1-z^2)^{m/2})

simplify

\frac{1}{2} ln (e^{m x} e^{m y} (1 - z^{2})^{\frac{m}{2}})

\frac{m x + m y}{2} + \frac{m ln ( 1 - z ^{2} )}{4}

Solution steps

\frac{1}{2} ln (e^{m x} e^{m y} (1 - z^{2})^{\frac{m}{2}})

Apply log rule:

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

\frac{1}{2} ln (e^{m x} e^{m y} (1 - z^{2})^{\frac{m}{2}}) = \frac{1}{2} ln (e^{m x}) + \frac{1}{2} ln (e^{m y}) + \frac{1}{2} ln ((1 - z^{2})^{\frac{m}{2}})

= \frac{1}{2} ln (e^{m x}) + \frac{1}{2} ln (e^{m y}) + \frac{1}{2} ln ((- z^{2} + 1)^{\frac{m}{2}})

Apply log rule:

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

ln (e^{m x}) = m x ln (e)

= \frac{1}{2} m x ln (e) + \frac{1}{2} ln (e^{m y}) + \frac{1}{2} ln ((1 - z^{2})^{\frac{m}{2}})

Apply log rule:

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

ln (e^{m y}) = m y ln (e)

= \frac{1}{2} m x ln (e) + \frac{1}{2} m y ln (e) + \frac{1}{2} ln ((1 - z^{2})^{\frac{m}{2}})

Apply log rule:

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

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

= \frac{1}{2} m x ln (e) + \frac{1}{2} m y ln (e) + \frac{1}{2} \cdot \frac{m}{2} ln (1 - z^{2})

\frac{1}{2} m x ln (e) = \frac{m x}{2}

\frac{1}{2} m y ln (e) = \frac{m y}{2}

\frac{1}{2} \cdot \frac{m}{2} ln (1 - z^{2}) = \frac{m ln ( 1 - z ^{2} )}{4}

= \frac{m x}{2} + \frac{m y}{2} + \frac{m ln ( - z ^{2} + 1 )}{4}

Combine the fractions

\frac{m x}{2} + \frac{m y}{2} : \frac{m x + m y}{2}

= \frac{m x + m y}{2} + \frac{m ln ( - z ^{2} + 1 )}{4}

= \frac{m x + m y}{2} + \frac{m ln ( 1 - z ^{2} )}{4}

lo g_{10} (\frac{x}{y ^{8}})

- ln (y) + ln (y + 1)

- 12 ln (8) + 12 ln (4)

- \frac{3}{4} lo g_{2} (\frac{3}{4}) - \frac{1}{4} lo g_{10} (x)

(\frac{1}{0.143}) ln (\frac{5 \cdot 1 0 ^{9}}{34000})

What is 1/2 ln(e^{mx}e^{my}(1-z^2)^{m/2}) ?
The solution to 1/2 ln(e^{mx}e^{my}(1-z^2)^{m/2}) is (mx+my)/2+(mln(1-z^2))/4