What is ln((b^{1/2})/(c(0.8-c))) ?

The solution to ln((b^{1/2})/(c(0.8-c))) is 1/2 ln(b)-ln(c)-ln(0.8-c)

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simplify ln((b^{1/2})/(c(0.8-c)))

simplify

ln (\frac{b ^{\frac{1}{2}}}{c ( 0.8 - c )})

\frac{1}{2} ln (b) - ln (c) - ln (0.8 - c)

Solution steps

ln (\frac{b ^{\frac{1}{2}}}{c ( 0.8 - c )})

Apply log rule:

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

ln (\frac{b ^{\frac{1}{2}}}{c ( 0.8 - c )}) = ln (b^{\frac{1}{2}}) - ln (c (0.8 - c))

= ln (b^{\frac{1}{2}}) - ln (c (- c + 0.8))

Apply log rule:

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

ln (b^{\frac{1}{2}}) = \frac{1}{2} ln (b)

= \frac{1}{2} ln (b) - ln (c (0.8 - c))

Apply log rule:

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

ln (c (0.8 - c)) = ln (c) + ln (0.8 - c)

= \frac{1}{2} ln (b) - (ln (c) + ln (- c + 0.8))

- (ln (c) + ln (0.8 - c)) : - ln (c) - ln (0.8 - c)

= \frac{1}{2} ln (b) - ln (c) - ln (0.8 - c)

ln (x^{3}) + ln (x)

\frac{1}{8} ln (\frac{17}{26})

ln (A) - ln (B) + 4 ln (C)

ln (\frac{112}{133})

lo g_{e} (4 x^{2} y^{5})

What is ln((b^{1/2})/(c(0.8-c))) ?
The solution to ln((b^{1/2})/(c(0.8-c))) is 1/2 ln(b)-ln(c)-ln(0.8-c)