drag the values to the correct locations on the image. not all values will be used\nfunction ( f ) is a…

drag the values to the correct locations on the image. not all values will be used\nfunction ( f ) is a logarithmic function with a vertical asymptote at ( x = 0 ) and an ( x )-intercept at ( (4,0) ). the function is decreasing over the interval ( (0, infty) ).\nfunction ( g ) is represented by the equation ( g(x)=log _{2}(x + 3)-2 ).\nover which interval are both functions positive?\n4 1 ( -infty ) 2 ( infty ) 0\n( (square, square) )

drag the values to the correct locations on the image. not all values will be used\nfunction ( f ) is a logarithmic function with a vertical asymptote at ( x = 0 ) and an ( x )-intercept at ( (4,0) ). the function is decreasing over the interval ( (0, infty) ).\nfunction ( g ) is represented by the equation ( g(x)=log _{2}(x + 3)-2 ).\nover which interval are both functions positive?\n4 1 ( -infty ) 2 ( infty ) 0\n( (square, square) )

Answer

Explanation:

Step1: Find the function ( f(x) )

Since ( f(x) ) is a logarithmic function with vertical asymptote ( x = 0 ), ( x )-intercept ( (4,0) ) and is decreasing, its general form is ( f(x)=\log_{\frac{1}{a}}x) ((a>1)). Using the (x)-intercept: when (y = 0), (x = 4), so (f(x)=\log_{\frac{1}{4}}x=-\log_{4}x). (f(x)>0) when (0 < x<4) (because the function is decreasing, (y>0) for (x) values between the vertical asymptote (x = 0) and the (x)-intercept (x = 4)).

Step2: Find when (g(x)>0)

Set (g(x)=\log_{2}(x + 3)-2>0). Add (2) to both sides: (\log_{2}(x + 3)>2). Using the property (y=\log_{a}u), if (y>k) ((a>1)), then (u>a^{k}). Here (a = 2), (k = 2), so (x+3>2^{2}), (x+3>4), (x>1). Also, the domain of (g(x)) is (x+3>0) (i.e., (x>-3)). So (g(x)>0) when (x>1).

Step3: Find the intersection of the two intervals

The interval for (f(x)>0) is ((0,4)) and for (g(x)>0) is ((1,\infty)). The intersection of ((0,4)) and ((1,\infty)) is ((1,4)).

Answer:

((1,4))