Answered By Leonardo F

I think the alternatives for this question are missing. The complete question is: 

Which statement is true for  $f\left(x\right)=\log_{\frac{1}{2}}x$ƒ (x)=log₁₂ x when  $x2>x1$x2>x1 ?

a)  $f\left(x1\right)>f\left(x2\right)$ƒ (x1)>ƒ (x2) 

b)  $f\left(x2\right)>f\left(x1\right)$ƒ (x2)>ƒ (x1) 

c)   $f\left(x1\right)>0;f\left(x2\right)<0$ƒ (x1)>0;ƒ (x2)<0  

d)  $f\left(x2\right)>0;f\left(x1\right)<0$ƒ (x2)>0;ƒ (x1)<0 

We have to start this exercise by taking a look at the following inequality:

 $x2>x1$x2>x1 

This means that when we calculate the logarithm, when we evaluate  $\log_{\frac{1}{2}}x2$log₁₂ x2 , we will always have a bigger argument of the logarithm (number that is inside the log) than when we evaluate  $\log_{\frac{1}{2}}x1$log_1‍2 x1 .

Since our base of the logarithm is less than 1, we have to power  $\frac{1}{2}$12  by a smaller number in the case of $\log_{\frac{1}{2}}x2$log₁₂ x2 than in the case of  $\log_{\frac{1}{2}}x1$log₁₂ x1 . We can test what was said in the last sentence by making up values of x1 and x2. Let's say that $x1=2$‍x1=2 and $x2=4$x2=4. Then, we have:

 $f\left(x1\right)=\log_{\frac{1}{2}}2=-1$ƒ (x1)=log₁₂ 2=−1 

 $f\left(x2\right)=\log_{\frac{1}{2}}4=-2$ƒ (x2)=log₁₂ 4=−2 

Hence, no matter what values we pick for x1 and x2, we will always have that the function evaluated at x1 is greater than the function evaluated at x2. Writing in mathematical symbols:

 $f\left(x1\right)>f\left(x2\right)$ƒ (x1)>ƒ (x2) 

Letter b will be the answer.

Answered By Leonardo F

Just a small correction: letter A is the answer.