What is ln(x)-4ln(8x^3+27)-3/2 ln(x-12) ?

The solution to ln(x)-4ln(8x^3+27)-3/2 ln(x-12) is ln(x/((8x^3+27)^4(x-12)^{3/2)})

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simplify ln(x)-4ln(8x^3+27)-3/2 ln(x-12)

simplify

ln (x) - 4 ln (8 x^{3} + 27) - \frac{3}{2} ln (x - 12)

ln (\frac{x}{( 8 x ^{3} + 27 ) ^{4} ( x - 12 ) ^{\frac{3}{2}}})

Solution steps

ln (x) - 4 ln (8 x^{3} + 27) - \frac{3}{2} ln (x - 12)

Apply log rule:

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

4 ln (8 x^{3} + 27) = ln ((8 x^{3} + 27)^{4})

= ln (x) - ln ((8 x^{3} + 27)^{4}) - \frac{3}{2} ln (x - 12)

Apply log rule:

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

ln (x) - ln ((8 x^{3} + 27)^{4}) = ln (\frac{x}{( 8 x ^{3} + 27 ) ^{4}})

= ln (\frac{x}{( 8 x ^{3} + 27 ) ^{4}}) - \frac{3}{2} ln (x - 12)

Apply log rule:

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

\frac{3}{2} ln (x - 12) = ln ((x - 12)^{\frac{3}{2}})

= ln (\frac{x}{( 8 x ^{3} + 27 ) ^{4}}) - ln ((x - 12)^{\frac{3}{2}})

Apply log rule:

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

ln (\frac{x}{( 8 x ^{3} + 27 ) ^{4}}) - ln ((x - 12)^{\frac{3}{2}}) = ln (\frac{\frac{x}{( 8 x ^{3} + 27 ) ^{4}}}{( x - 12 ) ^{\frac{3}{2}}})

= ln (\frac{\frac{x}{( 8 x ^{3} + 27 ) ^{4}}}{( x - 12 ) ^{\frac{3}{2}}})

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= ln (\frac{x}{( 8 x ^{3} + 27 ) ^{4} ( x - 12 ) ^{\frac{3}{2}}})

ln (\frac{5 a}{7 x + 6})

ln (\frac{s ^{2} + 25}{s ^{2} + 9})

ln (\frac{x ^{8} x - 6}{4 x + 5})

\frac{1}{2} lo g_{7} (4 m^{2})

lo g_{10} (\frac{7 x + 3}{x ^{4} ( x - 5 ) ^{2}})

What is ln(x)-4ln(8x^3+27)-3/2 ln(x-12) ?
The solution to ln(x)-4ln(8x^3+27)-3/2 ln(x-12) is ln(x/((8x^3+27)^4(x-12)^{3/2)})