What is expand 1/2 log_{3}(x)-2log_{3}(y+8) ?

The solution to expand 1/2 log_{3}(x)-2log_{3}(y+8) is log_{3}((x^{1/2})/(y^2+16y+64))

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expand 1/2 log_{3}(x)-2log_{3}(y+8)

expand

\frac{1}{2} lo g_{3} (x) - 2 lo g_{3} (y + 8)

lo g_{3} (\frac{x ^{\frac{1}{2}}}{y ^{2} + 16 y + 64})

Solution steps

\frac{1}{2} lo g_{3} (x) - 2 lo g_{3} (y + 8)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{2} lo g_{3} (x) = lo g_{3} (x^{\frac{1}{2}})

= lo g_{3} (x^{\frac{1}{2}}) - 2 lo g_{3} (y + 8)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 lo g_{3} (y + 8) = lo g_{3} ((y + 8)^{2})

= lo g_{3} (x^{\frac{1}{2}}) - lo g_{3} ((y + 8)^{2})

Apply log rule:

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

lo g_{3} (x^{\frac{1}{2}}) - lo g_{3} ((y + 8)^{2}) = lo g_{3} (\frac{x ^{\frac{1}{2}}}{( y + 8 ) ^{2}})

= lo g_{3} (\frac{x ^{\frac{1}{2}}}{( y + 8 ) ^{2}})

(y + 8)^{2} = y^{2} + 16 y + 64

= lo g_{3} (\frac{x ^{\frac{1}{2}}}{y ^{2} + 16 y + 64})

- lo g_{10} (3.9 \cdot 1 0^{- 13})

68 ln^{2} (\frac{49}{25})

2 lo g_{m} (w) - \frac{1}{3} lo g_{m} (x) + 3 lo g_{m} (z)

lo g_{10} (4) + 2 lo g_{10} (x)

What is expand 1/2 log_{3}(x)-2log_{3}(y+8) ?
The solution to expand 1/2 log_{3}(x)-2log_{3}(y+8) is log_{3}((x^{1/2})/(y^2+16y+64))