What is ln(1)-ln(7e^{-x}) ?

The solution to ln(1)-ln(7e^{-x}) is x-ln(7)

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simplify ln(1)-ln(7e^{-x})

simplify

ln (1) - ln (7 e^{- x})

x - ln (7)

Solution steps

ln (1) - ln (7 e^{- x})

Apply log rule:

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

ln (7 e^{- x}) = ln (7) + ln (e^{- x})

= ln (1) - (ln (e^{- x}) + ln (7))

Apply log rule:

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

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

= ln (1) - (ln (7) - x ln (e))

Apply log rule:

lo g_{a} (1) = 0

= 0 - (- ln (e) x + ln (7))

x ln (e) = x

= 0 - (- x + ln (7))

0 - (ln (7) - x) = - (ln (7) - x)

= - (- x + ln (7))

Distribute parentheses

= - (- x) - (ln (7))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= x - ln (7)

lo g_{5} (\frac{a ^{3}}{b ^{2}})

\frac{1}{2} (lo g_{6} (x) + lo g_{6} (y))

- lo g_{10} (3.3 \cdot 1 0^{- 6})

2 lo g_{a} (3) - 2 lo g_{a} (2)

ln (2) - ln (x)