What is ln(y^3)+1/3 ln(x^9y^{12})-7ln(y) ?

The solution to ln(y^3)+1/3 ln(x^9y^{12})-7ln(y) is ln(x^3)

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simplify ln(y^3)+1/3 ln(x^9y^{12})-7ln(y)

simplify

ln (y^{3}) + \frac{1}{3} ln (x^{9} y^{12}) - 7 ln (y)

ln (x^{3})

Solution steps

ln (y^{3}) + \frac{1}{3} ln (x^{9} y^{12}) - 7 ln (y)

Apply log rule:

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

\frac{1}{3} ln (y^{12} x^{9}) = ln ((y^{12} x^{9})^{\frac{1}{3}})

= ln (y^{3}) + ln ((y^{12} x^{9})^{\frac{1}{3}}) - 7 ln (y)

Apply log rule:

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

ln (y^{3}) + ln ((y^{12} x^{9})^{\frac{1}{3}}) = ln (y^{3} (y^{12} x^{9})^{\frac{1}{3}})

= ln (y^{3} (y^{12} x^{9})^{\frac{1}{3}}) - 7 ln (y)

Simplify

y^{3} (x^{9} y^{12})^{\frac{1}{3}} : y^{7} x^{3}

= ln (y^{7} x^{3}) - 7 ln (y)

Apply log rule:

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

7 ln (y) = ln (y^{7})

= ln (y^{7} x^{3}) - ln (y^{7})

Apply log rule:

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

ln (y^{7} x^{3}) - ln (y^{7}) = ln (\frac{y ^{7} x ^{3}}{y ^{7}})

= ln (\frac{y ^{7} x ^{3}}{y ^{7}})

Cancel the common factor:

y^{7}

= ln (x^{3})

ln (2 \cdot 4)

ln (\frac{1}{3}) - ln (\frac{1}{2})

ln (\frac{e ^{3 x} ( x + 1 )}{2 x ^{12}})

3 \cdot lo g_{10} (x - 3) - 4.5 lo g_{10} (x - 2)

\frac{ln ( \frac{30}{7} )}{3 ln ( 3 )}

What is ln(y^3)+1/3 ln(x^9y^{12})-7ln(y) ?
The solution to ln(y^3)+1/3 ln(x^9y^{12})-7ln(y) is ln(x^3)