What is 3log_{a}(5x^2)-1/4 log_{a}(8x+19) ?

The solution to 3log_{a}(5x^2)-1/4 log_{a}(8x+19) is log_{a}((125x^6(8x+19)^{3/4})/(8x+19))

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simplify 3log_{a}(5x^2)-1/4 log_{a}(8x+19)

simplify

3 lo g_{a} (5 x^{2}) - \frac{1}{4} lo g_{a} (8 x + 19)

lo g_{a} (\frac{125 x ^{6} ( 8 x + 19 ) ^{\frac{3}{4}}}{8 x + 19})

Solution steps

3 lo g_{a} (5 x^{2}) - \frac{1}{4} lo g_{a} (8 x + 19)

Apply log rule:

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

3 lo g_{a} (5 x^{2}) = lo g_{a} ((5 x^{2})^{3})

= lo g_{a} ((5 x^{2})^{3}) - \frac{1}{4} lo g_{a} (8 x + 19)

Apply log rule:

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

\frac{1}{4} lo g_{a} (8 x + 19) = lo g_{a} ((8 x + 19)^{\frac{1}{4}})

= lo g_{a} ((5 x^{2})^{3}) - lo g_{a} ((8 x + 19)^{\frac{1}{4}})

Apply log rule:

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

lo g_{a} ((5 x^{2})^{3}) - lo g_{a} ((8 x + 19)^{\frac{1}{4}}) = lo g_{a} (\frac{( 5 x ^{2} ) ^{3}}{( 8 x + 19 ) ^{\frac{1}{4}}})

= lo g_{a} (\frac{( 5 x ^{2} ) ^{3}}{( 8 x + 19 ) ^{\frac{1}{4}}})

Simplify

\frac{( 5 x ^{2} ) ^{3}}{( 8 x + 19 ) ^{\frac{1}{4}}} : \frac{125 x ^{6} ( 8 x + 19 ) ^{\frac{3}{4}}}{8 x + 19}

= lo g_{a} (\frac{125 x ^{6} ( 8 x + 19 ) ^{\frac{3}{4}}}{8 x + 19})

lo g_{10} ((5 x - 3)^{2}) + lo g_{10} ((2 x + 3)^{3})

[3 ln (3) - 1 - 2 ln (2)] - [3 ln (3) - 1 - 2 ln (2)]

ln (24) - ln (3)

ln (\frac{e ^{5}}{10})

lo g_{8} (4 x)

What is 3log_{a}(5x^2)-1/4 log_{a}(8x+19) ?
The solution to 3log_{a}(5x^2)-1/4 log_{a}(8x+19) is log_{a}((125x^6(8x+19)^{3/4})/(8x+19))