What is log_{3}(147)-log_{3}(70)-log_{3}(189) ?

The solution to log_{3}(147)-log_{3}(70)-log_{3}(189) is -log_{3}(90)

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simplify log_{3}(147)-log_{3}(70)-log_{3}(189)

simplify

lo g_{3} (147) - lo g_{3} (70) - lo g_{3} (189)

- lo g_{3} (90)

Decimal Notation

- 4.09590 \dots

Solution steps

lo g_{3} (147) - lo g_{3} (70) - lo g_{3} (189)

Apply log rule:

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

lo g_{3} (147) - lo g_{3} (70) = lo g_{3} (\frac{147}{70})

= lo g_{3} (\frac{147}{70}) - lo g_{3} (189)

Apply log rule:

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

lo g_{3} (\frac{147}{70}) - lo g_{3} (189) = lo g_{3} (\frac{\frac{147}{70}}{189})

= lo g_{3} (\frac{\frac{147}{70}}{189})

\frac{\frac{147}{70}}{189} = \frac{1}{90}

= lo g_{3} (\frac{1}{90})

\frac{1}{90} = 9 0^{- 1}

= lo g_{3} (9 0^{- 1})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), x > 0

lo g_{3} (9 0^{- 1}) = - 1 \cdot lo g_{3} (90)

= - 1 \cdot lo g_{3} (90)

Apply rule:

1 \cdot a = a

= - lo g_{3} (90)

ln (2 \cdot 1)

x - ln ∣ x + 4 ∣ + ln ∣ x ∣

\frac{4 \cdot π \cdot 1 0 ^{- 7}}{2 \cdot π} \cdot ln (\frac{27}{16})

5 ln (x^{3}) - 2 ln (x^{4})

\frac{1}{4} lo g_{c} (x) + 2 lo g_{c} (y) - 4 lo g_{c} (x)

What is log_{3}(147)-log_{3}(70)-log_{3}(189) ?
The solution to log_{3}(147)-log_{3}(70)-log_{3}(189) is -log_{3}(90)