What is ln(1/(e^2u)) ?

The solution to ln(1/(e^2u)) is -ln(u)-2

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simplify ln(1/(e^2u))

simplify

ln (\frac{1}{e ^{2} u})

- ln (u) - 2

Solution steps

ln (\frac{1}{e ^{2} u})

Apply log rule:

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

ln (\frac{1}{e ^{2} u}) = ln (1) - ln (e^{2} u)

= ln (1) - ln (e^{2} u)

Apply log rule:

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

ln (e^{2} u) = ln (e^{2}) + ln (u)

= ln (1) - (ln (u) + ln (e^{2}))

Apply log rule:

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

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

= ln (1) - (2 ln (e) + ln (u))

Apply log rule:

lo g_{a} (1) = 0

= 0 - (ln (u) + 2 ln (e))

2 ln (e) = 2

= 0 - (ln (u) + 2)

0 - (2 + ln (u)) = - (2 + ln (u))

= - (ln (u) + 2)

Distribute parentheses

= - (ln (u)) - (2)

Apply minus-plus rules

+ (- a) = - a

= - ln (u) - 2

ln (\frac{x ^{2} + 1}{x ^{3}})

2 ln (5 y)

- \frac{ln ( 3 )}{ln ( 3 ) - ln ( 2 )}

lo g_{e} (\frac{1}{3} e^{3 x + 1})

5 ln (2) - ln (3)