What is 10ln(e^{-5})-log_{3}(1/242) ?

The solution to 10ln(e^{-5})-log_{3}(1/242) is log_{3}(242)+5

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simplify 10ln(e^{-5})-log_{3}(1/242)

simplify

10 ln (e^{- 5}) - lo g_{3} (\frac{1}{242})

lo g_{3} (242) + 5

Decimal Notation

9.99624 \dots

Solution steps

10 ln (e^{- 5}) - lo g_{3} (\frac{1}{242})

Apply log rule:

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

lo g_{3} (\frac{1}{242}) = lo g_{3} (1) - lo g_{3} (242)

= 10 ln (e^{- 5}) - (lo g_{3} (1) - lo g_{3} (242))

Apply log rule:

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

ln (e^{- 5}) = - 5 ln (e)

= 10 - 5 ln (e) - (lo g_{3} (1) - lo g_{3} (242))

5 ln (e) = 5

- (lo g_{3} (1) - lo g_{3} (242)) = lo g_{3} (242)

= 10 - 5 + lo g_{3} (242)

Subtract the numbers:

10 - 5 = 5

= lo g_{3} (242) + 5

2 lo g_{10} (x - y) - lo g_{10} (x + y)

ln (- 5 e^{- 5 x})

lo g_{3} (b) + lo g_{3} (b + 4)

lo g_{10} (\frac{x}{100 y ^{7}})

ln (x^{5} 3 - x)

What is 10ln(e^{-5})-log_{3}(1/242) ?
The solution to 10ln(e^{-5})-log_{3}(1/242) is log_{3}(242)+5