What is 3/2*log_{3}(a)-1/4*log_{3}(16a^2) ?

The solution to 3/2*log_{3}(a)-1/4*log_{3}(16a^2) is log_{3}(a/2)

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simplify 3/2log_{3}(a)-1/4log_{3}(16a^2)

simplify

\frac{3}{2} \cdot lo g_{3} (a) - \frac{1}{4} \cdot lo g_{3} (16 a^{2})

lo g_{3} (\frac{a}{2})

Solution steps

\frac{3}{2} lo g_{3} (a) - \frac{1}{4} lo g_{3} (16 a^{2})

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

- \frac{1}{4} lo g_{3} (16 a^{2}) = - lo g_{3} ((16 a^{2})^{\frac{1}{4}})

= lo g_{3} (a^{\frac{3}{2}}) - lo g_{3} ((16 a^{2})^{\frac{1}{4}})

Apply log rule:

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

lo g_{3} (a^{\frac{3}{2}}) - lo g_{3} ((16 a^{2})^{\frac{1}{4}}) = lo g_{3} (\frac{a ^{\frac{3}{2}}}{( 16 a ^{2} ) ^{\frac{1}{4}}})

= lo g_{3} (\frac{a ^{\frac{3}{2}}}{( 16 a ^{2} ) ^{\frac{1}{4}}})

Simplify

\frac{a ^{\frac{3}{2}}}{( 16 a ^{2} ) ^{\frac{1}{4}}} : \frac{a}{2}

= lo g_{3} (\frac{a}{2})

ln (\frac{e ^{x}}{1 + e ^{x}})

lo g_{10} (\frac{5 x}{y})

ln (x) - 3 [ln (x - 7) + ln (x + 7)]

ln (\frac{e ^{5}}{9})

lo g_{b} (\frac{x ^{3} y}{z ^{5}})

What is 3/2*log_{3}(a)-1/4*log_{3}(16a^2) ?
The solution to 3/2*log_{3}(a)-1/4*log_{3}(16a^2) is log_{3}(a/2)