Respuesta :

27. (1/2*log(9) -(4*log(x) +1/3*log(3))/log(a)
  = ((1/2)*2*log(3) -4*log(x) -1/3*log(3))/log(a)
  = ((2/3)*log(3) -4*log(x))/log(a)
  = [tex]\frac{2}{3}\log_a{(3)}-4\log_a{(x)}[/tex]
[tex]\bf \textit{logarithm of factors} \\\\ log_a(xy)\implies log_a(x)+log_a(y) \\\\\\ \textit{Logarithm of rationals} \\\\ log_a\left( \frac{x}{y}\right)\implies log_a(x)-log_a(y) \\\\\\ \textit{Logarithm of exponentials} \\\\ log_a\left( x^b \right)\implies b\cdot log_a(x)\\\\ -------------------------------[/tex]

[tex]\bf log_a\left( \cfrac{9^{\frac{1}{2}}}{x^43^{\frac{1}{3}}} \right)\implies log_a\left( 9^{\frac{1}{2}} \right)-log_a\left( x^43^{\frac{1}{3}} \right) \\\\\\ log_a\left( 9^{\frac{1}{2}} \right)-\left[ log_a\left( x^4 \right)+log_a\left( 3^{\frac{1}{3}} \right) \right] \\\\\\ log_a\left( 9^{\frac{1}{2}} \right)-log_a\left( x^4 \right)-log_a\left( 3^{\frac{1}{3}} \right)\\\\\\ \cfrac{1}{2}log_a(9)-4log_a(x)-\cfrac{1}{3}log_a(3)[/tex]

or more expanded so as 

[tex]\bf \cfrac{1}{2}log_a(3^2)-4log_a(x)-\cfrac{1}{3}log_a(3) \\\\\\ \cfrac{1}{2}\cdot 2log_a(3)-4log_a(x)-\cfrac{1}{3}log_a(3) \\\\\\ log_a(3)-4log_a(x)-\cfrac{1}{3}log_a(3)[/tex]
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