What is log_{4}(e)-log_{4}(5e)-log_{4}(4e) ?

The solution to log_{4}(e)-log_{4}(5e)-log_{4}(4e) is -log_{4}(20e)

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simplify log_{4}(e)-log_{4}(5e)-log_{4}(4e)

simplify

lo g_{4} (e) - lo g_{4} (5 e) - lo g_{4} (4 e)

- lo g_{4} (20 e)

Decimal Notation

- 2.88231 \dots

Solution steps

lo g_{4} (e) - lo g_{4} (5 e) - lo g_{4} (4 e)

Apply log rule:

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

lo g_{4} (e) - lo g_{4} (5 e) = lo g_{4} (\frac{e}{5 e})

= lo g_{4} (\frac{e}{5 e}) - lo g_{4} (4 e)

Cancel the common factor:

e

= lo g_{4} (\frac{1}{5}) - lo g_{4} (4 e)

Apply log rule:

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

lo g_{4} (\frac{1}{5}) - lo g_{4} (4 e) = lo g_{4} (\frac{\frac{1}{5}}{4 e})

= lo g_{4} (\frac{\frac{1}{5}}{4 e})

\frac{\frac{1}{5}}{4 e} = \frac{1}{20 e}

= lo g_{4} (\frac{1}{20 e})

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - lo g_{4} (20 e)

2 (- ln ∣ x + 3 ∣ + 2 ln ∣ x + 6 ∣) + C

- \frac{1}{2} ln (x) + ln (\frac{1}{x})

ln (y + 4) - ln (x + 3)

ln (8 x^{2} - 4) - ln (x^{2})

\frac{8 ln ( 3 )}{2 ln ( 3 ) - ln ( 2 )}

What is log_{4}(e)-log_{4}(5e)-log_{4}(4e) ?
The solution to log_{4}(e)-log_{4}(5e)-log_{4}(4e) is -log_{4}(20e)