Respuesta :
The expression is [tex]\displaystyle{ \log_{\displaystyle{ \frac{1}{2}}}( \frac{3x^2}{2} )[/tex].
If the argument is a rational expression, like in our case: [tex]\displaystyle{ \frac{3x^2}{2} [/tex], we convert the expressions or numbers in the numerator into sums, and subtract the ones in the denominator.
For example: [tex]\log_b \displaystyle{ \frac{klm}{pq}= \log_k+\log_l+\log_m-\log_p-\log_q[/tex].
So, our logarithm can be expanded as [tex]\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} 3+\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} x^2-\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} 2[/tex].
We also have the rule [tex]\log_{ \displaystyle{ a^b}}c^d= \frac{d}{b}\log_a c [/tex].
Thus, using this rule we can perform the following simplifications:
[tex]\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} 3+\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} x^2-\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} 2=\displaystyle{ \log_{ \displaystyle {2^{-1}}}} 3^1+\displaystyle{ \log_{ \displaystyle {2^{-1}}}} x^2-\displaystyle{ \log_{ \displaystyle {2^{-1}}}}2^1[/tex]
[tex]=\frac{1}{-1}\log_2 3 +\frac{2}{-1}\log_2 x -\frac{1}{-1}\log_2 2=-\log_23-2\log_2x+1 [/tex]
Answer: [tex]1-\log_23-2\log_2x[/tex]
If the argument is a rational expression, like in our case: [tex]\displaystyle{ \frac{3x^2}{2} [/tex], we convert the expressions or numbers in the numerator into sums, and subtract the ones in the denominator.
For example: [tex]\log_b \displaystyle{ \frac{klm}{pq}= \log_k+\log_l+\log_m-\log_p-\log_q[/tex].
So, our logarithm can be expanded as [tex]\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} 3+\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} x^2-\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} 2[/tex].
We also have the rule [tex]\log_{ \displaystyle{ a^b}}c^d= \frac{d}{b}\log_a c [/tex].
Thus, using this rule we can perform the following simplifications:
[tex]\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} 3+\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} x^2-\displaystyle{ \log_{ \displaystyle {\frac{1}{2}}} 2=\displaystyle{ \log_{ \displaystyle {2^{-1}}}} 3^1+\displaystyle{ \log_{ \displaystyle {2^{-1}}}} x^2-\displaystyle{ \log_{ \displaystyle {2^{-1}}}}2^1[/tex]
[tex]=\frac{1}{-1}\log_2 3 +\frac{2}{-1}\log_2 x -\frac{1}{-1}\log_2 2=-\log_23-2\log_2x+1 [/tex]
Answer: [tex]1-\log_23-2\log_2x[/tex]
