What is 1/2 log_{b}(3)+3log_{b}(13)-3log_{b}(2) ?

The solution to 1/2 log_{b}(3)+3log_{b}(13)-3log_{b}(2) is log_{b}((2197*3^{1/2})/8)

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simplify 1/2 log_{b}(3)+3log_{b}(13)-3log_{b}(2)

simplify

\frac{1}{2} lo g_{b} (3) + 3 lo g_{b} (13) - 3 lo g_{b} (2)

lo g_{b} (\frac{2197 \cdot 3 ^{\frac{1}{2}}}{8})

Solution steps

\frac{1}{2} lo g_{b} (3) + 3 lo g_{b} (13) - 3 lo g_{b} (2)

Apply log rule:

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

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

= lo g_{b} (3^{\frac{1}{2}}) + 3 lo g_{b} (13) - 3 lo g_{b} (2)

Apply log rule:

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

3 lo g_{b} (13) = lo g_{b} (1 3^{3})

= lo g_{b} (3^{\frac{1}{2}}) + lo g_{b} (1 3^{3}) - 3 lo g_{b} (2)

Apply log rule:

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

lo g_{b} (3^{\frac{1}{2}}) + lo g_{b} (1 3^{3}) = lo g_{b} (1 3^{3} \cdot 3^{\frac{1}{2}})

= lo g_{b} (1 3^{3} \cdot 3^{\frac{1}{2}}) - 3 lo g_{b} (2)

1 3^{3} = 2197

= lo g_{b} (2197 \cdot 3^{\frac{1}{2}}) - 3 lo g_{b} (2)

Apply log rule:

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

3 lo g_{b} (2) = lo g_{b} (2^{3})

= lo g_{b} (2197 \cdot 3^{\frac{1}{2}}) - lo g_{b} (2^{3})

Apply log rule:

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

lo g_{b} (2197 \cdot 3^{\frac{1}{2}}) - lo g_{b} (2^{3}) = lo g_{b} (\frac{2197 \cdot 3 ^{\frac{1}{2}}}{2 ^{3}})

= lo g_{b} (\frac{2197 \cdot 3 ^{\frac{1}{2}}}{2 ^{3}})

2^{3} = 8

= lo g_{b} (\frac{2197 \cdot 3 ^{\frac{1}{2}}}{8})

ln (u - 2) - ln (u + 2)

\frac{lo g _{3} ( a ^{2} b )}{lo g _{3} ( z ^{3} )}

50 ln (10) - 50 ln (5) - 25

lo g_{3} (x y^{2})

30 - 12 ln (x) - 18 ln (y)

What is 1/2 log_{b}(3)+3log_{b}(13)-3log_{b}(2) ?
The solution to 1/2 log_{b}(3)+3log_{b}(13)-3log_{b}(2) is log_{b}((2197*3^{1/2})/8)