What is 2/3 log_{b}(x+1)-1/4 log_{b}(x+2) ?

The solution to 2/3 log_{b}(x+1)-1/4 log_{b}(x+2) is log_{b}(((x+1)^{2/3}(x+2)^{3/4})/(x+2))

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simplify 2/3 log_{b}(x+1)-1/4 log_{b}(x+2)

simplify

\frac{2}{3} lo g_{b} (x + 1) - \frac{1}{4} lo g_{b} (x + 2)

lo g_{b} (\frac{( x + 1 ) ^{\frac{2}{3}} ( x + 2 ) ^{\frac{3}{4}}}{x + 2})

Solution steps

\frac{2}{3} lo g_{b} (x + 1) - \frac{1}{4} lo g_{b} (x + 2)

Apply log rule:

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

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

= lo g_{b} ((x + 1)^{\frac{2}{3}}) - \frac{1}{4} lo g_{b} (x + 2)

Apply log rule:

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

\frac{1}{4} lo g_{b} (x + 2) = lo g_{b} ((x + 2)^{\frac{1}{4}})

= lo g_{b} ((x + 1)^{\frac{2}{3}}) - lo g_{b} ((x + 2)^{\frac{1}{4}})

Apply log rule:

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

lo g_{b} ((x + 1)^{\frac{2}{3}}) - lo g_{b} ((x + 2)^{\frac{1}{4}}) = lo g_{b} (\frac{( x + 1 ) ^{\frac{2}{3}}}{( x + 2 ) ^{\frac{1}{4}}})

= lo g_{b} (\frac{( x + 1 ) ^{\frac{2}{3}}}{( x + 2 ) ^{\frac{1}{4}}})

Simplify

\frac{( x + 1 ) ^{\frac{2}{3}}}{( x + 2 ) ^{\frac{1}{4}}} : \frac{( x + 1 ) ^{\frac{2}{3}} ( x + 2 ) ^{\frac{3}{4}}}{x + 2}

= lo g_{b} (\frac{( x + 1 ) ^{\frac{2}{3}} ( x + 2 ) ^{\frac{3}{4}}}{x + 2})

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8 ln 2 + 4 - 4 + 2 ln 2 + 4 + 1

What is 2/3 log_{b}(x+1)-1/4 log_{b}(x+2) ?
The solution to 2/3 log_{b}(x+1)-1/4 log_{b}(x+2) is log_{b}(((x+1)^{2/3}(x+2)^{3/4})/(x+2))