What is ln(1/(2xy^2)) ?

The solution to ln(1/(2xy^2)) is -ln(x)-2ln(y)-ln(2)

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simplify ln(1/(2xy^2))

simplify

ln (\frac{1}{2 x y ^{2}})

- ln (x) - 2 ln (y) - ln (2)

Solution steps

ln (\frac{1}{2 x y ^{2}})

Apply log rule:

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

ln (\frac{1}{2 x y ^{2}}) = ln (1) - ln (2 x y^{2})

= ln (1) - ln (2 x y^{2})

Apply log rule:

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

ln (2 x y^{2}) = ln (2) + ln (x) + ln (y^{2})

= ln (1) - (ln (x) + ln (y^{2}) + ln (2))

Apply log rule:

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

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

= ln (1) - (ln (2) + ln (x) + 2 ln (y))

Apply log rule:

lo g_{a} (1) = 0

= 0 - (ln (x) + 2 ln (y) + ln (2))

0 - (ln (2) + ln (x) + 2 ln (y)) = - (ln (2) + ln (x) + 2 ln (y))

= - (ln (x) + 2 ln (y) + ln (2))

Distribute parentheses

= - (ln (x)) - (2 ln (y)) - (ln (2))

Apply minus-plus rules

+ (- a) = - a

= - ln (x) - 2 ln (y) - ln (2)

ln (\frac{x}{12})

ln (70) - ln (40)

ln (\frac{8}{e})

ln (25) - ln (14)

- ln (x + 2) + ln (x)