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dedylja [7]
3 years ago
7

Expand the log as a sun or difference

Mathematics
2 answers:
Hitman42 [59]3 years ago
5 0
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)
  = \frac{2}{3}\log_a{(3)}-4\log_a{(x)}
snow_lady [41]3 years ago
4 0
\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)\\\\
-------------------------------

\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)

or more expanded so as 

\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)
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