Question

drag the values to the correct locations on the image. not all values will be used
function ( 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) ).
function ( g ) is represented by the equation ( g(x)=log _{2}(x + 3)-2 ).
over which interval are both functions positive?
4 1 ( -infty ) 2 ( infty ) 0
( (square, square) )

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:

\((1,4)\)